Examinable Question and answer system programming

FACULTY OF COMPUTING AND INFORMATION TECHNOLOGY

SYSTEMS PROGRAMMING CSC 2102

TEST 1 Marking Guide

EXAM 2006

A collection of programs whose purpose is to make more effective use of the

computer.

Examples include language processors (compilers and assemblers), loaders,

macro processors, operating systems and file

Programs written by the user for the purpose of solving particular problems

using a computer as a tool e.g. application packages.

Decode the instructions and use them to control the activities within the

system. It also performs the arithmetic and logical computations.

Stores both data and instructions that are currently being used.

� Decode the address of the memory location being accessed.

� Buffer data onto/off the bus.

� Contain circuitry to perform memory reads or write.

1. Immediate Addressing: Information is part of the instruction.

2. Direct addressing: The address is part of the instruction.

3. Register addressing: The operand is in the register and the register’s address

is part of the instruction.

4. Indirect Addressing: The address is in the location whose address is

specified as part of the instruction.

5. Base addressing: The address is formed by adding the contents of a memory

location or register to a number called a displacement which is part of the

instruction.

6. Indexing: It is a process of incrementing or decrementing an address as the

computer sequences through a set of consecutive or evenly spaced addresses.

7. Auto incrementing / decrementing: The index is automatically incremented

by an instruction.

Instructions: They are translated into machine code by the assembler.

e.g. MOV A , M

Directive: Gives directions to the assembler during the assembly process but

they are not translated into machine code. e.g. the directive COST : DS 1

will causes the assembler to reserve a byte and associate a symbol COST to

it

Reserves memory and assigns a label to the first byte of the reserved area.

e.g. ARRAY: DS 20 reserves 20 bytes and assigns the label ARRAY to

the byte with the lowest address.

DB (Define Byte)

Puts values into or pre-assign values to memory locations as well as reserve

space and assign labels. e.g. DATA: DB 5 , puts a value 5 into a reserved

memory location called DATA.

DW (Define Word)

Puts words into or pre-assign values to memory locations as well as reserve

space and assign labels.

RESW Reserve the indicated number of words for a data area.

RESB Reserve the indicated number of bytes for a data area

WORD Generate a one word integer constant

BYTE Generate character or hexadecimal constant, occupying as

many bytes as needed to represent the constant

A Accumulator; used for arithmetic Operations

X Index register; used for addressing

L Linkage register; the Jump to subroutine (JSUB) instruction stores the

return address in this register.

PC Program Counter; Contains the address of the next instruction to be fetched

for execution.

SW Status word; contains the condition codes

B Base register, used for addressing

S General Working register, no special use.

T General Working register, no special use.

F Floating Point Accumulator.

Loads immediate Data 5 in the Accumulator. Addressing

Mode is Immediate

Loads immediate Data 6 in register B. Addressing Mode is

Immediate

Adds contents of register B to contents in register A

Addressing Mode is Register

Stores contents of register A to an address ABCDE

Addressing Mode is Simple

The program loads registers A and B with values 5 and 6 respectively. It

adds the contents of both registers and stores the answer a memory location

whose address is ABCDE

LDA ZERO

STA SUM

LDX SUM

LOOP: LDA SUM

ADD NUMS, X

STA SUM

LDA INDEX

ADD THREE

STA INDEX

COMP K300

JLT LOOP

INDEX RESW 1

RESULT RESW 1

SUM RESW 1

NUMS RESW 100

ZERO WORD 0

K300 WORD 300



TEST 2

TEST 2007

Answer all questions

Time allowed: 1 Hour

1. (a) Explain the following (i) SYMTAB

A symbol table; used to hold addresses of symbols used in the

program.

(ii) LOCCTR A Location Counter: used to help in the assignment of addresses. It is

initialized to the beginning address specified in the START statement.

After each source statement is processed the length of the assembled

instruction or data area to be generated is added to the LOCCTR.

(iii) OPTAB Operation Code Table: used to look up and validate operation codes in

the source program. In pass 2 it is used to translate the opcodes to

machine language

(iv) LITTAB Literal table: It contains the literal name, the operand value and the

address assigned to the literal

(b) Two passes are usually used to assemble an Assembler language

program; explain briefly what is accomplished in each pass. Pass 1

(i) The first pass scans the source program for label definitions and

assigns addresses to all statements in the program.

(ii) Saves the addresses assigned to all labels for use in pass 2

(iii) Perform some processing of the assembler directives. e.g., determining

the length of data areas defined by BYTE, RESW etc.

Pass 2 (assemble the instructions and generate machine code)

(i) Assemble instructions (translating operation codes and looking up

addresses)

(ii) Generate data values defined by BYTE, WORD, etc.

(iii) Perform processing of the directives not done during pass 1.

(iv) Write the object program and assembly listing onto some output device

which will later be loaded in memory for execution.

2. Given below is a program PROG that has been written using the SIC/XE

machine. LDT NUM LDA ALPHA ADDR T, A SUB #8 STA SUM1 LDA GAMMA ADDR T,A SUB #12 STA SUM2

ALPHA RESW 1 SUM1 RESW 1 GAMMA RESW 1 SUM2 RESW 1 NUM RESW1 1

(i) Study the program carefully and write mathematical expressions

that correspond to the given code. SUM1 = ALPHA + NUM – 8

SUM2 = GAMMA + NUM - 12

(ii) Assuming that it is to be loaded in memory from 0058H, generate

the corresponding addresses for each instruction and write the corresponding machine code. The operation codes for the instructions are LDT = 74, LDA = 00, ADDR = 90, SUB = 1C, STA = 0C and the addresses for registers A and T are 0 and 5 respectively.

0058 LDT NUM 772022

005B LDA ALPHA 032013

005E ADDR T, A 9050

0060 SUB #8 1D0008

0063 STA SUM1 0F200E

0066 LDA GAMMA 03200E

0069 ADDR T, A 9050

006B SUB #12 1D000C

006E STA SUM2 0F2003

0071 ALPHA RESW 1

0074 SUM1 RESW 1

0077 GAMMA RESW 1

007A SUM2 RESW 1

007D NUM RESW 1

(iii) Write the corresponding text listing for the generated machine instructions. H^ PROG ^000058^000028

T^000058^19^772022^032013^9050^1D0008^0F200E^03200E^9050^1D000C^0F2003

E^000058

3. (a) (i) What is program relocation? Ability of a program to be able to run successfully in memory without

considering the address where it is going to be loaded.

(ii) Explain why some instruction need modification records

during program relocation and others do not. Instruction that have been assembled using addresses are the ones that

need to be modified; format 1, format 2 do need have address

operands; and format 3 instructions that have been assembled using

program counter relative or base relative are similarly not affected

because the displacements do not change when the program is loaded

at different addresses.

(b) Assume the following instruction

0006 LDA NUMBER 2032 NUMBER: WORD 6

(i) Write machine code for the instruction at address 0006 03102032

(ii) Write the modification record for the instruction

M00000705

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System software are programs that control the operation of the machine and carry out the

most basic functions the computer performs. They control the way in which the computer

receives input, produces output, manages and stores data, carries out or executes

instructions of other programs etc. Examples of systems programs include language

processors (compilers and assemblers that accept people like languages and translate them

into machine language), loaders (they prepare machine language programs for execution),

macro processors (allow programmers to use abbreviations), operating systems and file

systems (allow flexible storing and retrieval of information).

Application programs are written by the user for the purpose of solving particular problems

using a computer as a tool e.g. application packages like MSWORD, EXCEL etc.�� & �� #�'��#�#��!��'��"��#�� #�� "��

The microprocessor

• Decodes the instructions and use them to control the activities within the system

• Performs the arithmetic( + , -, /, *) and logical (>,>=,<,<=, =, =!) computations.

Timing Circuitry (Clock)

Used to synchronise the activities within the microprocessor and the bus control logic.

Memory

Stores both data and instructions that are currently being used. locations.

I/O Devices

Devices for communicating with the external world.

Mass Storage Devices.

Computer components for permanently storing programs and data.

System Bus.

A set of conductors that connect the CPU to its memory and I/O devices. The bus conductors

are normally separated into 3 groups:

• The Data Lines: for transmitting information

• Address Lines: Indicate where information is to come from or where it is to be

placed.

• Control Lines: To regulate the activities on the bus.

Interface

Circuitry needed to connect the bus to a device

�(')� & �� #�� #� �� #� '��#�#�� !� �� '�� '"��"�'��"��The control unit consisting of:

(i) The Program Counter (PC): It holds the address of the main memory location from

which the next instruction is to be fetched.

(ii) Instruction Register (IR) Receives the instruction when it is brought from memory

and holds it while it gets decoded and executed.

(iii) Processor Status Word (PSW) contains condition flags which indicate the current

status of the CPU and the important characteristics of the result of the previous

instruction.

(iv) Stack Pointer (SP): Accesses a special part of memory called a stack. It hold the

address at the top of the stack.

Working Registers

(i) Arithmetic Registers: They temporarily hold the operands and the result of the

arithmetic operations.

(ii) Address Registers: for addressing data and instructions in main memory.

Arithmetic/Logic Unit

It performs arithmetic and logical operations on the contents of the working registers, the

PC, memory locations etc.. It also sets and clears the appropriate flags.

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The format of an instruction is the arrangement of the instruction with respect to assigning

meaning to its various groups of bits is called its format.

It consists of

Operation code: The portion of the instruction that specifies what the instruction

Operands: Any address or piece of data that is required by the instruction to complete its

execution

e.g.

Register to register transfer

Opcode Destination Source

(Register B) (Register A)

($) �� #��#��"��#��*�*+��"�� #,��-�8. Immediate Addressing: Information is part of the instruction. No addressing is

needed to get the information.

e.g. X = 3 + 5 (3 and 5 are constatnts)

9. Direct addressing: The address is part of the instruction.

e.g. X = p + q (where p and q are memory locations)

10. Register addressing: The operand is in the register and the register’s address is part

of the instruction.

e.g. X = A + B (where A and B are registers)

11. Indirect Addressing: The address is in the location whose address is specified as part

of the instruction. This location may be a register (register indirect addressing) or it

may be a memory location.

e.g. add contents of register R1 to the memory location whose address is in register

R2.

12. Base addressing: The address is formed by adding the contents of a memory location

or register to a number called a displacement which is part of the instruction. It is

used primarily to reference arrays or in relocating a program in memory.

e.g X = 5 + B:5 where B is a base register

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��01��'� #�� SIC SIC/XE

Memory (215

) (220

)

Registers A,X, L, PC, SW Has extra registers namely B, S, T, F

Data Formats No floating point hardware Has a floating point format

0 1 0 0 0 1 1 1

1 11 36

Instruction

Formats

Only one Instruction Format 8 1 15

4 different instruction formats 8

8 4 4

6 1 1 1 1 1 1 12

6 1 1 1 1 1 1 20

Addressing

Modes

Direct and Indexed only Immediate, Indirect, Base Relative,

Program Counter relative, Simple,

Direct

Instruction set Fewer Instructions More Instructions

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#��"�'� �#� X PC B

2050 3000 4000

Memory addresses

Opcode n i x b p e displacement

6FAC00 = 011011 11 1 01 0 110000000000

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50 1400 2000

3000 2000 1000

Opcode x address

S exponent fraction

Opcode

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Opcode n i x b p e address

S exponent fraction

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TEST 2

1. Explain the use of the following:

(ii) The OPTAB ��

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(3 marks)�

(iii) The SYMTAB �� (3 marks)

(iv) The LOCCTR

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(v) The LITTAB �� (3 marks)

2. Assume the following assembler code for the SIC/XE machine

Study the code and briefly explain what is being accomplished

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(3 marks)

(ii) Generate machine code for the assembler code.

(3 marks each instruction = 27

marks)

(iii) Write the object code for this program as it would appear on

an output device. %&'#�(&))))))&))))*+��&))))))&))))!,&-.)))$&*/))!+&)/))))&)$�)!0&!1�)$!&)2�)0+&,)0!&�)

!/&$1"22)�+&))))))��

(5 marks)

3. (a) What are literals? ��3��4 5+�25��$��

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(3 marks)

Explain the difference between literals and immediate

operands. Use an

example to explain your answer

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4. (a) What is Program Relocation and how is it done? �� 3�� 3��3��3��3��8�� 9��:� �� (3 marks)��

(b) Consider the code below

Write all the Modification Records that may be needed in this

piece of code. Why don’t the other remaining instructions need

modification records? ;��#��3�� )))- �))!$��))"-��;��)))-��;)))))*)/�;��))!$��;))))!0)/�;�� ))"-� �� ;))))"*)/��(6 marks)� ��.��7$�3��

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Assume that RDREC and WRREC are indifferent

control sections; rewrite the machine code for the

instructions at addresses 0006, 0013 and 0026� � � � � � � � � � � � � � � � � � � �

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� � � � � � � � � � � � � � � � � � � � � (6 marks)

(ii) Modify the program below so that it can run on a

machine that uses only direct addressing like the simple

SIC machine. ))))� �'<� ��#�� )� �

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(24 marks)

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(i) An option that allows the selection of alternative sources of input e.g.

INCLUDE program_name (library_name) directs the loader to read the

designated object program from a library and treat it as if it were part of the

primary loader input.

(ii) An option to allow the user to delete external symbols or entire control

sections e.g. DELETE csect_name instructs the loader to delete the named

control section from the set of programs being loaded.

(iii) An option to change external symbols e.g. CHANGE name1, name2 causes

the external symbol name1 to be changed to name2 wherever it appears in the

object programs.

(iv) An option to specify alternative libraries to be searched e.g. LIBRARY

MYLIB will cause library MYLIB to be searched before the standard system

libraries.

(v) Loaders that perform automatic library search may be asked to exclude some

functions that come with the library search e.g. NOCALL STDEV, PLOT,

CORREL will instruct the loader to exclude the said functions. This avoids the

overhead of loading and linking the unneeded routines and saves on memory

space.

(vi) Options to control outputs from the loader. The user can specify whether an

output is needed or not.

(vii) Options to specify whether external references will be resolved by library

search.

(viii) An option to specify the location at which execution is to begin thus overriding

any information given in the object program.

(ix) An option to control whether or not the loader should attempt to execute the

program if errors are detected during the load.

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Segments of the program that can execute in such a way that segments that are

not needed in memory at any particular time do not get loaded��

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Segment Starting Address

Relative Actual Length

1 0000 5000 1000

2 1000 6000 1800

3 2800 7800 4000

4 2800 7800 3000

5 5800 A800 1000

6 1000 6000 400

7 1400 6400 1800

8 2C00 7C00 800

9 1400 6400 2000

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execution domain and

(ii) To provide diagnostics for violations of the programming language

semantics in the source program�

A

B

C D/E

F

G

H/I

J

K

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(i) A grammar

The grammar specifies the form or syntax of legal statements in the

language.

(ii) Lexical Analysis

The task of scanning the source statement, to recognize and classify the

various

(iii) Syntactic Analysis

Recognizing each statement in the program as some language

construct, such as a declaration, or an assignment statement,

described by the grammar

(iv) A token

Tokens are the fundamental building blocks of the language. It might

be a keyword, a variable name, an integer, an arithmetic operator

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<exp>

<exp> <term>

<term>

<term> <factor>

<factor>

<factor>

id id id

{ALPHA} - {BETA} * {GAMMA}

(ii) ALPHA DIV (BETA + GAMMA) – DELTA <expr>

<expr> <term>

<term>

<term> <factor>

<expr> <factor>

<factor>

<expr>

<term> <term>

<factor> <factor>

id DIV id + id - id

{ALPHA} {BETA} {GAMMA} {DELTA} �

OR

<term>

<term> <factor>

<factor> <expr>

>

<expr>

<term>

<expr> <term>

<term>

<factor> <factor>

<factor>

id DIV id + id -

id

{ALPHA} {BETA} {GAMMA}

{DELTA} �

(�) �� "�� #�� !�"� �� ,"��"�%�"��#��%�"��"��'��% �� #,��"��<exp> ::= <factor> | <exp> + <factor> | <exp> - factor> | <exp> * <factor>

| <exp> DIV <factor>

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<factor> <expr>

<expr>

<factor>

<expr> <factor>

<term>

<factor>

id DIV id + id -

id

{ALPHA} {BETA} {GAMMA}

{DELTA} ��

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ALPHA DIV (BETA + GAMMA) - DELTA; Id1 DIV ( id2 + id3 ) - id4 ;

MAKERERE UNIVERSITY

INSTITUTE OF COMPUTER SCIENCE

B.SC.CSC 2004/2005 ACADEMIC YEAR

YEAR TWO, SEMESTER I

FINAL EXAMINATION

PAPER CSC2102: SYSTEMS PROGRAMMING

Marking Guide

DATE: - 27TH

JAN 2005

TIME: - 2:00PM - 5:00 P.M.

DURATION: - 3 HOURS

INSTRUCTIONS

Candidates MUST answer ALL questions in section A and any THREE questions from Section B

Section A: (40 marks): All Questions are Compulsory.

1. Use examples to help you differentiate between Systems Software and Application

Software.

System software are programs that control the operation of the machine and carry out

the most basic functions the computer performs. They control the way in which the

computer receives input, produces output, manages and stores data, carries out or

executes instructions of other programs etc. Examples of systems programs include

language processors (compilers and assemblers that accept people like languages

and translate them into machine language), loaders (they prepare machine language

programs for execution), macro processors (allow programmers to use

abbreviations), operating systems and file systems (allow flexible storing and

retrieval of information).

Application programs are written by the user for the purpose of solving particular

problems using a computer as a tool e.g. application packages like MSWORD,

EXCEL etc. (4 Marks)�

2. Use an example to help you explain the Format of the Machine Language

Instruction

The format of an instruction is the arrangement of the instruction with respect to

assigning meaning to its various groups of bits. It consists of

Operation code: The portion of the instruction that specifies what the instruction

Operands: Any address or piece of data that is required by the instruction to complete

its execution

e.g.

Register to register transfer

Opcode Destination Source

(Register B) (Register A) (4 Marks)

3. (a) What are the main functions of an Assembler?

An assembler is a program that accepts as input an assembler language

program and it produces its machine language equivalent along with

information for the loader (2 Marks)

(b) Explain the steps involved in assembling a simple computer program.

� Convert mnemonic operation codes to their machine language

equivalent; e.g. translate STL to 14.

� Convert symbolic operands to their equivalent machine addresses e.g.

translate RETADR to 1033.

� Build the machine instructions in the correct format.

� Convert the data constants into their internal machine representations

e.g. translate EOF to 454F46

� Write the object program and the assembly listing. (4 Marks)

0 1 0 0 0 1 1 1

4. 6FAC00 and 064050 are two machine language instructions for the SIC/XE

machine. The contents of registers X, PC, and B are shown and a snapshot of

some memory locations with their contents is also shown. Explain

(v) What each instruction accomplishes.

(vi) The addressing modes used.

(vii) The target addresses where applicable.

(8 ) Contents of the registers involved after execution of each instruction.� �

X PC B

2050 3000 4000

Memory addresses


6FAC00 = 011011 11 1 01 0 110000000000

(iv) Opcode = 011011 = 6C = LDS = Load register S from memory

(v) Addressing mode = Simple (Direct) addressing (because n = 1 and i = 1)

(vi) Target address : x = 1, p = 1, displacement = C00

50 + 1400 + C00 = 2050

(iv) Contents of Register S = 3000 (4 Marks)


064050 = 000001 1 0 0 10 0 000001010000

(v) Opcode = 000001 = 04 = LDX = Load register X from memory

(vi) Addressing mode = Indirect addressing (because n = 1 and i = 0)

(vii) Target address :, b = 1, displacement = 050

2000 + 050 = 2050

(iv) Contents of Register X = 2000 (4 Marks)

5. Use relevant examples to help you explain the difference between the following:

�� Program Blocks and Control Sections

Program Blocks are segments of code that are re-arranged within a single object unit

while Control Sections are parts of the program that maintain their identities after

assembly; each section can be loaded and relocated independently of each other.

references in the usual way because the assembler does not know where any other

control section will be loaded at execution time. (2 Marks)

Relevant Examples (2 Marks)

�� EXTDEFS and EXTREFS statements

The EXTDEF statement names symbols called external symbols that are defined in

this control section but may also be used by other sections.

EXTREF names symbols used in this section but are defined elsewhere. (2 Marks)

Relevant Examples (2 Marks)

6. (a) What is LTORG?

50 1400 2000

3000 2000 1000

It is a directive used to indicate where the literal pool has been placed in a

program if they are not put at the bottom of the program. (2 Marks)

(b) Explain what happens when an assembler encounters the LTORG

directive.

It creates a literal pool that contains all the literal operands used since the

previous LTORG or the beginning of the program. (2 Marks)

8. What are the main functions of a compiler?

• To generate code to implement meaning of a source program in the execution

domain and

• To provide diagnostics for violations of the programming language semantics

in the source program. (2 Marks)

9. Explain the following:

(i) A grammar

The grammar specifies the form or syntax of legal statements in the language.

(1 Mark)

(ii) Lexical Analysis

The task of scanning the source statement, to recognize and classify the

various tokens (1 Mark)

(iii) Syntactic Analysis

Recognizing each statement in the program as some language construct, such

as a declaration, or an assignment statement, described by the grammar

(1 Mark)

(iv) A token

Tokens are the fundamental building blocks of the language. It might be a

keyword, a variable name, an integer, an arithmetic operator (1 Mark)

Section B: (60 marks): Attempt only THREE questions from this section.

9. (a) List and explain the main components of a typical microprocessor

• The Program Counter (PC): It holds the address of the main memory location

from which the next instruction is to be fetched. (1 Mark)

• Instruction Register (IR) Receives the instruction when it is brought from

memory and holds it while it gets decoded and executed. (1 Mark)

• Processor Status Word (PSW) contains condition flags which indicate the

current status of the CPU and the important characteristics of the result of the

previous instruction. (1 Mark)

• Stack Pointer (SP): Accesses a special part of memory called a stack. It hold

the address at the top of the stack. (1 Mark)

• Arithmetic Registers: They temporarily hold the operands and the result of

the arithmetic operations. (1 Mark)

• Address Registers: for addressing data and instructions in main memory.

(1 Mark)

• Arithmetic/Logic Unit: It performs arithmetic and logical operations on the

contents of the working registers, the PC, memory locations etc. It also sets

and clears the appropriate flags. (1 Mark)

(b) List and state the uses of the registers found in the SIC/XE

microprocessor.

A Accumulator: used for arithmetic Operations

X Index register: used for addressing

L Linkage register: the Jump to subroutine (JSUB) instruction stores the return

address in this register.

PC Program Counter: Contains the address of the next instruction to be fetched

for execution.

SW Status word: contains the condition codes

B Base register: used for addressing

S General Working register: no special use.

T General Working register: no special use.

F Floating Point Accumulator: For floating point arithmetic (1/2 mark each)

(b) List all the registers of the Intel 8085 microprocessor that have similar

functions like those ones listed in part (b) above.

SIC/XE Intel 8085

Accumulator (A) Accumulator

Linkage Register (L) Stack Pointer

Program Counter (PC) Program Counter (PC)

Status Word (SW) Processor Status Word (PSW)

General Working Registers

(S, T)

General Working Registers

(B, C, D, E, H, L)

(4 Marks)

(c) Explain the formats of the machine language instructions for the SIC/XE

computer.

Format 1 (1 byte)

8

Format 2 (2 bytes)

8 4 4

Format 3 (3 bytes)

6 1 1 1 1 1 1 12

Format 4 (4 bytes)

6 1 1 1 1 1 1 20

(4 Marks)

10. (a) An array ARR1 at a memory location 0030 contains 10 integers. Write a

SIC/XE Assembler Language program that will add up all the integers of

this array and store the answer at a memory location GREATER if the

answer is greater than 100 or at a memory location LESS if the answer is

less than 100

LDS #3

LDT #30

LDX #0

LDA ARR1, X

LOOP : ADDR S, X

ADD ARR1, X

Opcode

opcode r1 r2

opcode n i x b p e disp

opcode n i x b p e address

COMPR X, T

JLT LOOP

COMP #100

JLT SMALLER

STA GREATER

J OUT

SMALLER STA LESS

OUT

ARR1 RESW 10

GREATER RESW 1

LESS RESW 1 (10 Marks)

(b) Assume that the program is to be assembled from memory location 000H

and that GREATER is at a memory location after the array ARR1 and

LESS is at a memory location next to GREATER generate machine code

for this program using Appendix C

Addresses Label Instruction Operand Machine Code

0000 ADDINT START 0

0000 LDS #3 6D0003

0003 LDT #30 75001E

0006 LDX #0 050000

0009 LDA ALPHA, X 03A014

000C LOOP: ADDR S ,X 9041

000E ADD ALPHA,X 1BA00F

0011 COMPR X,T A015

0013 JLT LOOP 3B2FF6

0016 COMP #100 290064

0019 JLT SMALLER 3B2006

001C STA GREATER 0F202F

001F J OUT 3F2003

0022 SMALLER STA LESS 0F202A

0025 OUT

0030 ARR1: RESW 10

004E GREATER: RESW 1

004F LESS RESW 1

(10 Marks)

11. (a) Explain the following terms used in Systems Programs:

(i) Assembling

Produce machine code and Assembler Listing from an assembler

language. (1 Mark)

(ii) Loading

bring the object program into memory for execution. (1 Mark)

(iii) Relocation

modify the object program so that it can be loaded at an address

different from the location originally specified. (1 Mark)

Linking

Combine two or more separate object programs and supplies the

information needed to allow references between them (1 Mark)

(b) Briefly explain one of the methods that may be used to assemble a

program in one pass.

One method produces object code directly in memory for immediate execution.

In this method no object program is written out and no loader is needed (load

and go). It avoids the overhead of writing the object program out and reading

it back in. The assembler generates object code instructions as it scans the

source program. If the operand is a symbol that has not yet been defined, the

operand address is omitted when the instruction is assembled. The symbol

used as an operand is entered in the symbol table. This entry is flagged as

undefined. When the definition for the symbol is encountered the symbol table

is scanned and the proper address is inserted into any instructions previously

generated. Any SYMTAB entries that are still undefined at the end of the

program are flagged by the assembler as errors.

The second type of a one pass assembler produces object code as output is

needed on systems where external working storage devices are not available.

Forward references are entered into lists as before but when the definition of

the symbol is encountered another text record with the correct operand

address is generated. When the program is loaded, this address will be

inserted into the instruction by the action of the loader. (5 Marks)

(c) (i) What are Overlay programs?

They are segments of the program that can execute in such a way that

segments that are not needed in memory at any particular time do not

get loaded�� (1 Mark)

(ii) The structure below shows 14 different control sections making up

an overlay structure. The number of bytes each control section

uses is also indicated within the particular section. Write

statements which can be used to create this structure.

SEGMENT SEG(A) PARENT SEG(A)

A

1000

H

900

C

500

B

2000

D/E

1000

G

700

I

1000

F

800

M

200

N

2000

J

2500

K/L

1000

O

850

P

900

SEGMENT SEG(B) SEGMENT SEG(H)

SEGMENT SEG(C) SEGMENT SEG(K/L)

SEGMENT SEG4(D/E) SEGMENT SEG(O)

SEGMENT SEG(F) SEGMENT SEG(P)

PARENT SEG(C) PARENT SEG(H)

SEGMENT SEG(M SEGMENT SEG(J)

SEGMENT SEG(N) PARENT SEG(A)

PARENT SEG(A) SEGMENT SEG(I)

SEGMENT SEG(G) (5 Marks)

(iii) Assuming the overlay structure of the control sections above,

complete the following table for all the control sections shown.



A 0000 2500 1000

B 1000 3500 2000

C 3000 5500 500

D/E 3500 5A00 1000

F 4500 6A00 800

M 3500 5A00 200

N 3700 5C00 2000

G 1000 3500 700

H 1000 3500 900

J 1900 3E00 2500

K/L 1900 3E00 1000

O 2900 4E00 850

P 3150 5650 900

I 1000 3500 1000

(5 Marks)

12 (a) The program below consists of three different control sections; write

down the machine code for the instructions and directives for control

section PROGA.

0000 PROGA START 0 MACHINE CODE EXTDEF LISTA,ENDA EXTREF LISTB, ENDB,LISTC,ENDC . 0020 REF1 LDA LISTA 03201D 0023 REF2 +LDT LISTB + 4 77100004 0027 REF3 LDX #ENDA - LISTA 050014 0040 LISTA EQU * 0054 ENDA EQU * 0054 REF4 WORD LISTC 000000 0057 REF5 WORD LISTB - 3 FFFFFD 005A REF6 WORD LISTA + LISTB 000040 005D REF7 WORD ENDC-LISTC- 100 FFFF9C 0060 REF8 WORD LISTA–LISTB–ENDA+ENDB FFFFEC END

(5 Marks)

(b) Write down the object code for PROGA as it would appear on an output

device.

H^PROGA ^000000^000063

D^LISTA ^000040ÊNDA ^000054

R^LISTB ÊNDB ^LISTC ÊNDC

T^000020^0A^03201D^77100004^050014

T^000054^0F^000000^FFFFFD^000040^FFFF9C^FFFFEC

M^000024^05^+LISTB

M^000054^06^+LISTC

M^000057^06^+LISTB

M^00005A^06^+LISTB

M^00005A^06^+PROGA

M^00005D^06^+ENDC

M^00005D^06^-LISTC

M^000060^06^-LISTB

M^000060^06^+ENDB

E^000020 (5 Marks)

(c) Draw an external symbol table (ESTAB) for this program assuming that

PROGA is to be loaded from a memory location 2500 and that PROGB

immediately follows PROGA and PROGC immediately follows PROGB.

Control Section Symbol Name Address Length

PROG A 2500 0063

LISTA 2540

ENDA 2554

PROGB 2563 007F

LISTB 25C3

ENDB 25D3

PROGC 25E2 0051

LISTC 2612

ENDC 2624

(4 Marks)

(d) Using the assumptions made in part (c) above to calculate the values for

REF4, REF5 and REF6

REF4 = LISTC = 2612 (2 Marks)

REF5 = LISTB – 3 = 25C3 – 3 = 25C0 (2 Marks)

REF6 = LISTA + LISTB = 2540 + 25C3 = 4B03 (2 Marks)

13 (a) Draw top down parse trees according to the grammar in Appendix A for

the following expressions:

(i) ALPHA – BETA * GAMMA

<exp>

<exp> <term>

<term>

<term> <factor>

<factor>

<factor>

id id id

{ALPHA} - {BETA} * {GAMMA} (5 marks)

(ii) ALPHA DIV (BETA + GAMMA) – DELTA

<expr>

<expr> <term>

<term>

<term> <factor> <factor>

<factor>

<expr>

<expr> <term>

<term>

<factor> <factor>

id DIV ( id + id ) - id

{ALPHA} {BETA} {GAMMA} {DELTA}

(5 marks)

(b) Use a bottom up method to parse the following statements using the

operator precedence table in Appendix B. Assume that a semicolon

preceded the given expressions.

(i) SUMQ := SUMQ + VALUE * VALUE;

(ii) ANS:= ALPHA DIV (BETA + GAMMA) – DELTA ;

SUMQ := SUMQ + VALUE * VALUE; Id1 := id2 + id3 * id4 ; <N1>

< = < >

id2

{SUMQ}

Id1 := <N1> + id3 * id4 ; <N2>

< = < < N2 >

id3

{VALUE}

Id1 := <N1> + <N2> * id4 ; <N3>

< = < < < >

id4

{VALUE}

Id1 := <N1> + <N2> * <N3> ; <N4>

< = < < N4 >

<N2> * <N3>

id1 := <N1> + <N4> ; <N5>

< = < N5 >

<N1> + <N4>

id1 := <N5> ; <N6>

< N6 >

id1 := <N5>

<N6>

<N5>

<N1> <N4>

<N2> <N3>

id1 := id2 + id3 * id4

{SUMQ} {SUMQ} {VALUE} {VALUE}

(10 marks)

ANS:= ALPHA DIV (BETA + GAMMA) – DELTA ; id1 := id2 DIV ( id3 + id4 ) - id5 ; <N1>

< = < N1 >

id2

{ALPHA}

id1 := <N1> DIV ( id3 + id4 ) - id5 ; <N2>

< = < < < N2 >

id3

{BETA}

id1 := <N1> DIV ( <N2> + <N3> ) - id5 ; <N3>

< = < < < N4 >

id4

{GAMMA}

id1 := <N1> DIV ( <N4> ) - id5 ; <N4>

< = < < N5 >

<N2> + <N3>

id1 := <N1> DIV <N5> - id5 ; <N5>

< = < N6 >

( <N4> )

id1 := <N6> - id5 ; <N6>

< = < < N7 > <N7>

id1 := <N6> - <N7> ; id5

< = < N8 > <N1> DIV <N5> {DELTA}

<N8> <N9>

id1 := <N8> ;

< = <N9> >

<N6> - <N7> id1 := <N8>

<N9>

<N8>

<N6> <N7>

<N1> <N5>

<N4>

<N2> <N3>

id1 := id2 DIV ( id3 + id4 ) - id5

{ANS} {ALPHA} {BETA} {GAMMA} {DELTA}

�� 9�1��1. <prog> ::= PROGRAM <prog-name> VAR <dec-list> BEGIN <stmt-list> END

2. <prog-name>::= id

3. <dec-list> ::= <dec> | <dec-list> ; <dec>

4. <dec> ::= <id-list> : <type>

5. <type> ::= INTEGER

6. <id-list> ::= id | <id-list> , id

7. <stmt-list> ::= <stmt> | <stmt-list> ; <stmt>

8. <stmt> ::= <assign> | <read> | <write> | for

9. <assign> ::= id := <exp>

10. <exp> ::= <term> | <exp> + <term> | <exp> - <term>

11. <term> ::= <factor> | <term * <factor> | <term> DIV <factor>

12. <factor> ::= id | int | ( <exp>)

13. <read> ::= READ ( <id-list> )

14. <write> ::= WRITE ( <id-list> )

15. <for> ::= FOR <index-exp> DO <body>

16. <index-exp> ::= id := <exp> TO <exp>

17. <body> ::= <stmt> | BEGIN <stmt-list> END

��

�� 9�1�6��

VAR BEGIN END INTEGER FOR READ WRITE TO DO ; : , := + - * DIV ( ) id int

PROGRAM = < VAR = < < < < BEGIN = < < < < < END > > INTEGER > = > < FOR READ = WRITE = TO > < < < < < < < DO < > < < < > <

; > > < < < > < < <

: > < >

, =

:= > = > < < < < < < <

+ > > > > > > < < < > < <

- > > > > > > < < < > < <

* > > > > > > > > < > < <

DIV > > > > > > > > < > < <

( > < < < < < < = < <

) > > > > > > > > >

Id > > > > > > > = > > > > >

Int > > > > > > > > >

APPENDIX C

SIC/XE INSTRUCTION SET

MNEMONIC FORMAT OPCODE EFFECT Notes

ADD m 3/4 18 A (A) + (m...m+2)

ADDF m 3/4 58 F (F) + (m...m+5) X F

ADDR r1,r2 2 90 r2 (r2) + (r1) X

AND m 3/4 40 A (A) & (m…m+2)

CLEAR r1 2 B4 r1 0 X

COMP m 3/4 28 (A) : (m…m+2) C

COMPF m 3/4 88 (F) : (m…m+5) X F C

COMPR r1,r2 2 A0 (r1) : (r2) X C

DIV m 3/4 24 A (A) / (m….m+2)

DIVF m 3/4 64 F (F) / (m….m+5) X F

DIVR r1,r2 2 9C r2 (r2) / (r1) X

FIX 1 C4 A (F) convert to integer X F

FLOAT 1 C0 F (A) convert to float X F

HIO 1 F4 Halt I/O channel number (A) P X

J m 3/4 3C PC m

JEQ m 3/4 30 PC m if CC set to =

JGT m 3/4 34 PC m if CC set to >

JLT m 3/4 38 PC m if CC set to <

JSUB m 3/4 48 L (PC); PC m

LDA m 3/4 00 A (m..m+2)

LDB m 3/4 68 B (m..m+2) X

LDCH m 3/4 50 A [rightmost byte] (m)

LDF m 3/4 70 F (m..m+5) X F

LDL m 3/4 08 L (m..m+2)

LDS m 3/4 6C S (m..m+2) X

LDT m 3/4 74 T (m..m+2) X

LDX m 3/4 04 X (m..m+2)

LPS m 3/4 D0 Load processor status from P X

MUL m 3/4 20 A (A) * (m..m+2)

MULF m 3/4 60 F (A) * (m..m+5) X F

MULR r1,r2 2 98 r2 (r2) * (r1) X

NORM 1 C8 F (F) normalised X F

OR m 3/4 44 A (A) | (m..m+2)

RD m 3/4 D8 A [rightmost byte] data from P

RMO r1,r2 2 AC r2 (r1) X

RSUB 3/4 4C PC (L)

SHIFTL r1,n 2 A4 r1 (r1) left shift n bits X

SHIFTR r1,n 2 A8 r1 (r1) right shift n bits X

SIO 1 F0 Start I/O channel number (A) P X

SSK m 3/4 EC Protection key for address m P X

STA m 3/4 0C m..m+2 (A)

STB m 3/4 78 m..m+2 (B) X

STCH m 3/4 54 m (A) [rightmost byte]

STF m 3/4 80 m..m+5 (F) X F

STI m 3/4 D4 Interval timer value (m ..m+2) P X

STL m 3/4 14 m..m+2 (L)

STS m 3/4 7C m..m+2 (S) X

STSW m 3/4 E8 m..m+2 (SW) P

STT m 3/4 84 m..m+2 (T) X

STX m 3/4 10 m..m+2 (X)

SUB m 3/4 1C A (A) - (m..m+2)

SUBF m 3/4 5C F (F) - (m..m+5) X F

SUBR r1,r2 2 94 r2 (r2) – (r1) X

SVC n 2 B0 Generate SVC Interrupt X

TD m 3/4 E0 Test Device specified by (m) P C

TIO 1 F8 Test I/O channel number (A) P X C

TIX m 3/4 2C X (X) + 1; (X) : (m..m+2) C

TIXR r1 2 B8 X (X) + 1; (X) : (r1) X C

WD m 3/4 DC Device specified by (m) (A) P

MAKERERE UNIVERSITY

FACULTY OF COMPUTING & INFORMATION TECHNOLOGY

END OF SEMESTER I EXAMINATION 2005/2006

PROGRAMME: Bachelor of Science in Computer Science

YEAR OF STUDY: Year II

COURSE NAME: Systems Programming

COURSE CODE: CSC 2102

DATE: 1st December 2005 TIME: 4.00p.m.-7.00 p.m.


1. (a) What is Systems Software?

A collection of programs whose purpose is to make more effective use of the

computer by controlling the operation of the machine and carry out the most

basic functions the computer performs. They control the way in which the

computer receives input, produces output, manages and stores data, carries

out or executes instructions of other programs etc (2 marks)

(b) List any three Systems software and explain their functions.

(i) Compilers: Translate high level languages to machine language

(ii) Assemblers: Translate Assembler languages to machine language

(iii) Loader: Prepare machine language programs for execution

(iv) Macro processors: allow programmers to use abbreviations

(v) Operating systems and file systems allow flexible storing and retrieval

of information. (3 marks)

2. (a) Explain the following:

(i) Operation Code

The portion of the instruction that specifies what the instruction does.

(1 mark)

(ii) Operand

Any address or piece of data that is required by the instruction to

complete its execution (1 mark)

(iii) Destination

The location that is changed or where the source operand is taken

(1 mark)

(b) Use examples to help you explain the difference between an instruction

and a Directive

An instruction is translated into machine code by the assembler

e.g. LDA NUM (1 mark)

A Directive gives directions to the assembler during the assembly process but

is not translated into machine code e.g. COST : DS 1 (1 mark)

3. How will the following be stored using the SIC/XE machine?

(i) A real number - 15.25

-15.25 = - 1111.01 = -0.111101 x 24

Exponent = 1024 + 4 = 1028 = 10000000100

1 10000000100 111101 = CO4F40000000 (2 marks)

(b) An integer -15

To be changed to 2’s complement.

15 = 1111 = 000000000000000000001111

= 111111111111111111110001 = FFFFF1 (2 marks)

4. (a) What are the basic functions of an assembler?

It accepts as input an assembler language program and it produces its

machine language equivalent along with information for the loader

(2 marks)

(b) Give the tables usually needed in assembling a machine language

program and state their uses.

(i) Symbol Table (SYMTAB)

A table used to store addresses assigned to labels. (1½ marks)

(ii) Operation Code Table (OPTAB)

A table used to look up mnemonic operation codes and translate them

into their machine language equivalents. (1½ marks)

(iii) Literal Table(LITTAB)

It contains the literal name, the operand value and the address

assigned to the operand when it is placed in the literal pool. (1 mark)

5. Assume the following SIC/XE code

0016 LDA @Num1

0050 Num1 BYTE 0100H


(i) What does the instruction at address 0016 do and what is the addressing

mode used?

It loads 0050H into the accumulator. Addressing Mode is Indirect Addressing

(2 marks)

(ii) Write machine code for this instruction.

Displacement = 0050 – 0019 = 037


000000 1 0 0 0 1 0 0 3 7 = 022037 (3 marks)

6. (a) What is Program Relocation?

Ability of loading a program anywhere in memory without stating the actual

starting address (1 mark)

(b) Assume the following SIC/XE instruction and write a modification

records for it. 0016 +LDA Number

M000001705 (2 marks)

7. (a) How are literals different from immediate operands?

With immediate addressing the operand value is assembled as part of the

machine instruction. With a literal the assembler generates the specified value

as a constant at some other memory location. The address of this generated

constant is used as the target address for the machine instruction. (2 marks)

(b) What is the difference between Absolute expressions and Relative

expressions?

An absolute expression contains only absolute terms or relative terms so long

as the relative terms occur in pairs and the terms in each pair have opposite

signs. A relative expression is one in which all the relative terms except one

can be paired. The remaining unpaired relative term must have a positive.

(2 marks)

8. (a) What is the basic function of a loader?

It brings an object program into memory and it start executing it. (1 mark)

(b) Explain the following abbreviations

(i) ESTAB

External symbol table: It is used to store names and addresses of each

external symbol in the set of control sections being loaded. It also

indicates in which control section the symbol is defined. (1 mark)

(ii) PROGADDR

It is the beginning address in memory where the linked program is loaded. Its

value is supplied to the loader by the operating system. (1 mark)

(iii) CSADDRR

CSADDR is the starting address assigned to the control section currently

being scanned by the loader. (1 mark)

9. (a) Explain the following:

(i) A Grammar

The grammar specifies the form or syntax of legal statements in the

language. (1 mark)

(ii) A Token

The fundamental building blocks of the language (1 mark)

(b) State the difference between the SYNTAX and SEMANTICS of a

grammar.

The syntax is the description of the form of legal statements in the language

while the Semantics is the meaning of the statements (2 marks)


10. (a) List and explain the uses of the main components of a typical computer

The microprocessor

Decode the instructions and use them to control the activities within the

system. It also performs the arithmetic and logical

Timing Circuitry (Clock)

Used to synchronise the activities within the microprocessor and the bus

control logic.

Memory

Stores both data and instructions that are currently being used. address.

The I/O Devices

Devices for communicating with the external world

Mass Storage Devices

Used for storing large quantities of information.

System Bus.

A set of conductors that connect the CPU to its memory and I/O devices.

Interfaces

Circuitry needed to connect the bus to a device.

Bus Control Unit

Co-ordinates the CPU activities with the external world. (14 marks)

(b) List and explain the uses of the registers found in the control unit of a

typical microprocessor.

The Program Counter (PC)

It holds the address of the main memory location from which the next

instruction is to be fetched.

Instruction Register (IR)

Receives the instruction when it is brought from memory and holds it while it

gets decoded and executed.

Processor Status Word (PSW)

Contains condition flags which indicate the current status of the CPU and the

important characteristics of the result of the previous instruction.

Stack Pointer (SP):

Accesses a special part of memory called a stack. It is used to temporarily

store important information while sub routines are being executed. It hold the

address at the top of the stack (6 marks)

11. (a) An array ARR1 contains 20 words of both positive and negative integers.

Write a SIC/XE program that will add up all these integers and store the

sum at a memory location POSIT if the sum is positive or NEGAT if the

sum is negative.

LDS #3

LDS #60

LDX #0

STX SUM

LOOP: LDA NUMBER, X

ADD SUM

STA SUM

ADDR S, X

COMPR X, T

JLT LOOP

LDS #0

COMPR A, S

JLT NEGATIVE

STA POSIT

J OUT

NEGATIVE: STA NEGAT

OUT: (10 marks)

(b) Generate machine code for the first five executable instructions of your

program assuming that the program is loaded from an address 0000H,

the beginning address of ARR1 is 1000H, memory location POSIT is at a

memory location 2000H and NEGAT is at a memory location 3000H

0000 ADD START 0000 0000 FIRST LDS #3 6D0003 0003 LDT #60 75003C 0006 LDX #0 050000 0009 STX SUM 132027 000C LOOP LDA ARR1, X 03AFF1 000F ADD SUM 1B2021 0012 STA SUM 0F201E 0015 ADDR S ,X 0018 COMPR X, T 001B JLT LOOP 001E LDS #0 0021 COMPR A , S 0024 JLT NEGATIVE 0027 STA POSIT 002A J OUT 002D NEGATIVE STA NEGAT 0030 OUT 0033 SUM RESW 1 1000 ARR1 RESW 20 2000 POSIT RESW 1 3000 NEGAT RESW 1

(10 marks)

12. (a) An Assembler Language program is assembled usually in two passes.

Why do we need these two passes? Explain briefly what is done in each

pass.

Addresses of the labels are not known.

Pass 1 (Define symbols)

• Scans the source program for label definitions and assigns addresses

to all statements in the program.

• Saves the addresses assigned to all labels for use in pass 2

• Perform some processing of the assembler directives. e.g., determining

the length of data areas defined by BYTE, RESW etc.

Pass 2 (assemble the instructions and generate machine code)

• Assemble instructions (translating operation codes and looking up

addresses

• Generate data values defined by BYTE, WORD, etc.

• Perform processing of the directives not done during pass 1.

• Write the object program and assembly listing onto some output device

which will later be loaded in memory for execution. (5

marks)

Slightly modify the above piece of code so that it can be assembled in one

pass. 1000 COPY START 1000 1000 EOF BYTE C’EOF’ 454F46 1003 THREE WORD 3 000003 1006 ZERO WORD 0 000000 1009 RETADR RESW 1 100C LENGTH RESW 1 100F BUFFER RESB 4096 200F FIRST STL RETADR 141009 2012 CLOOP JSUB RDREC 480000 2015 LDA LENGTH 00100C 2018 COMP ZERO 281006 201B JEQ ENDFIL 300000 201E JSUB WRREC 480000 2021 J CLOOP 3C2012 2024 ENDFIL LDA EOF 001000 2027 STA BUFFER 0C100F 202A LDA THREE 001003 202D STA LENGTH 0C100C 2030 JSUB WRREC 480000 2033 LDL RETADR 081009 2036 RSUB 4C0000

(5 marks)

Generate the text listing for the above code assuming a one pass

assembler that generates object code on an output device.

H^COPY ^001000^001039

T^001000^09^454F46^000003^000000

T^00200F^15^141009^480000^00100C^281006^300000^480000^3C2012

T^00201C^02^2024

T^002024^19^001000^0C100F^001003^0C100C^480000^081009^4C0000

(10 marks)

13. (a) List and explain any five common loader options.

(i) An option that allows the selection of alternative sources of input e.g.

INCLUDE program_name (library_name) directs the loader to read the

designated object program from a library and treat it as if it were part of

the primary loader input.

(ii) An option to allow the user to delete external symbols or entire control

sections e.g. DELETE csect_name instructs the loader to delete the

named control section from the set of programs being loaded.

(iii)An option to change external symbols e.g. CHANGE name1, name2

causes the external symbol name1 to be changed to name2 wherever it

appears in the object programs.

(iv) An option to specify alternative libraries to be searched e.g. LIBRARY

MYLIB will cause library MYLIB to be searched before the standard

system libraries.

(v) Loaders that perform automatic library search may be asked to exclude

some functions that come with the library search e.g. NOCALL STDEV,

PLOT, CORREL will instruct the loader to exclude the said functions. This

avoids the overhead of loading and linking the unneeded routines and

saves on memory space.

(vi) Options to control outputs from the loader. The user can specify whether

an output is needed or not.

(vii) Options to specify whether external references will be resolved by

library search.

(viii) An option to specify the location at which execution is to begin thus

overriding any information given in the object program.

(ix) An option to control whether or not the loader should attempt to execute

the program if errors are detected during the load. (5 marks)

( b) Given below is piece of code for a control section. There are also other

control sections in memory referencing this section. Assume that the

variables LENGTH and BUFFER of this section are referenced by other

control sections.

(i) Rewrite the piece of code above to reflect these assumptions.

0000 COPY START 0 EXTDEF LENGTH, EXTREF RDREC, 0000 FIRST STL RETADR 17202D 0003 LDB #LENGTH 69202D BASE LENGTH 0006 CLOOP +JSUB RDREC 4B100000 000A LDA LENGTH 032026 000D COMP #0 290000 0010 JEQ ENDFIL 332007 0013 +JSUB WRREC 4B100000 0017 J CLOOP 3F2FEC 001A ENDFIL LDA EOF 032010 001D STA BUFFER 0F2016 0020 LDA #3 010003 0023 STA LENGTH 0F200D 0026 +JSUB WRREC 4B100000 002A J @RETADR 3E2003 002D EOF BYTE C’EOF’ 454F46 0030 RETADR RESW 1 0033 LENGTH RESW 1 0036 BUFFER RESB 4096

(5 marks)

(ii) Generate a text listing for this code.

H^COPY ^000000^000039

D^LENGTH^000033^BUFFER^000036

R^RDREC ^WRREC

T^000000^1D^17202D^69202D^4B100000^032026^290000^332007^4B100000^3F

2FEC^032010

T^00001D^13^0F2016^010003^0F200A^4B100000^3E2000^454F46

M^000007^05^+RDREC

M^000014^05^+WRREC

M^000027^05^+WRREC

E^000000 (10

marks)

14. (a) What are programs with Overlay Structures?

They are segments of a program that can execute in such a way that

segments that are not needed in memory at any particular time do not get

loaded in memory. (3 marks)

(b) Assume program segments defined by the following Overlay structure.

SEGMENT SEG1(A)

SEGMENT SEG2(B)

SEGMENT SEG3(C)

PARENT SEG2

SEGMENT SEG4(D/K)

PARENT SEG1

SEGMENT SEG5(E)

SEGMET SEG6(F)

PARENT SEG5

SEGMENT SEG7(G)

PARENT SEG1

SEGMENT SEG8(H/L)

SEGMENT SEG9(I)

PARENT SEG8

SEGMENT SEG10(J)

i. Express this structure diagrammatically

(5 marks)

A

B

C D/K

E

F G

H/L

I J

(ii) Which segments can be in memory at the same time together with

segment J.

Segments A , H and L (2 marks)

(iii) Assume the lengths of the segments are as follows:

Segment Length Segment Length

(bytes) (bytes)

A 1000 G 400

B 1800 H 800

C 4000 I 1000

D 2800 J 800

E 800 K 2000

F 1000 L 2000

Assume also that the actual starting address of the program is 5000 fill

the following table below.



1 0000 5000 1000

2 1000 6000 1800

3 2800 7800 4000

4 2800 7800 4800

5 1000 6000 800

6 1800 6800 1000

7 1800 6800 400

8 1000 6000 2800

9 3800 8800 1000

10 3800 8800 800

(10 marks)



TEST 1

A study of the functions and structures of the programs used by the

system

A collection of programs whose purpose is to make more effective

use of the compute by controling the operation of the machine and

carry out the most basic functions the computer performs. They

control the way in which the computer receives input, produces

output, manages and stores data, carries out or executes instructions

of other programs etc

(i) Compilers: Translate high level languages to machine language

(ii) Assemblers: Translate Assembler languages to machine language

(iii) Loader: Prepare machine language programs for execution

(iv) Macro processors: allow programmers to use abbreviations

(v) Operating systems and file systems allow flexible storing and

retrieval of information.

b p Addressing Mode Target Address

0 0 Direct TA = Disp + (X)

0 1 PC Relative TA = (PC) + Disp + (X)

1 0 Base Relative TA = (B) + Disp + (X)

1 1 -----

n i Addressing Mode Operand

0 0 Simple Addressing Operand = Contents of TA (bits b,p,e are part of address bits)

0 1 Immediate Operand = TA

1 0 Indirect Addressing Word at TA is the address of the Operand

1 1 Simple addressing Operand = Contents of TA

Opcode R1 R2

Opcode n i x b p e Displacement

Opcode n i x b p e address

Explanation

Examples

Property SIC SIC/XE Memory 215 bytes 220 bytes

Registers 5 ( A, X, L, PC,W) Extra registers (B, S, T, F)

Instruction Formats Only 1 Format 4 different Formats

Addressing Modes Only 2 (Direct, Index) 5 (Immediate, Indirect, Direct, PC and Base

Data Formats No Floating Point Format Has Floating Point Format

(b)

.

SIC Intel 8085 A Accumulator

L Stack Pointer

PC Program Counter

SW Processor Status Word

� � � �

-21.25 = -10101.01 = -0.1010101 x 25

Exponent = 1024 + 5 = 1029 = 10000000101

=> 1 10000000101 101010100000000

= C 0 5 A A 0 0 0 0 0 0 0

= C05AA0000000

� � � �

- 15 = -1111 = -000000001111

invert bits and add 1 = 111111110000 + 1 = 111111110001

+- � � � �.�� 42��

��?�� "�� +�� +�� #�� , �,�� + #�� -�+� �� 84��#��?��#��+��# ��$��-��+��+��#��-�+�,��+��40��0� ��$� 6� �+�� ?�� $� #�� -� �� +�� #�� ?�� +�#��+��?�� B��#��+ #��##�� $$+��,��$�� $��#��+ #�� +,��$$+��-�+��#��+ #�� B��$��-��+�,+��!�

X PC B

50 90 A0

50 100 300

Memory addresses

opcode n i x b p e displacement

� �.��:��

00 = LDA Target address = b + x + Disp = A0 + 50 + 10 = 100

Loads 300 in register A

Addressing Mode = Direct (Simple)

Target Address = 100


� ��:��

18 = ADD m Target address = b + x + Disp = A0 + 50 + 10 = 100

Adds 200 to Register A = 500

Addressing Mode = Indirect

Target Address = 100


� �42��:��

0C = STA m Target address = b + Disp = A0 + 120 = 1C0

Stores 500 from the accumulator to a memory location 1C0

Addressing Mode = Direct

Target Address = 1C0



TEST 2

1. (a) List and explain all the activities that take place when an Assembler

language program is assembled.

• Convert mnemonic operation codes to their machine language

equivalent.

• Convert symbolic operands to their equivalent machine addresses

• Build the machine instructions in the correct format.

• Convert the data constants into their internal machine representations

• Write the object program and the assembly listing.

(b) Explain the following:

(i) SYMTAB

A table used to store addresses assigned to labels.

(ii) LOCCTR

A location counter LOCCTR is used to help in the assignment of

addresses. It is initialized to the beginning address specified in the

START statement. After each source statement is processed the length

of the assembled instruction or data area to be generated is added to

the LOCCTR

(iii) OPTAB

100 300 200

A table used to look up mnemonic operation codes and translate them

into their machine language equivalents.

2. Assume the following SIC/XE code



Explain what happens when each of the following instructions is executed. Write

machine code for each of the instructions and in each case state the addressing

mode and the format of the instruction which has been used.

(i) 0016 LDA NUM1

• Loads a number 0100H into the accumulator.

• Displacement = 0050 – 0019 = 037


000000 1 1 0 0 1 0 0 3 7 = 032037

Machine Code = 032037

• Addressing Mode: = Simple / Direct

• Format: Format 3

(ii) 0016 LDA @Num1

• Loads a number 0050H into the accumulator.

• Displacement = 0050 – 0019 = 037


000000 1 0 0 0 1 0 0 3 7 = 022037

Opcode = 022037

• Addressing Mode: = Indirect


(iii) 0016 LDA #5

• Loads a base ten number 5 into the accumulator.

• Displacement = 005


000000 0 1 0 0 0 0 0 0 5 = 010005

Machine code = 010005

• Addressing Mode: Immediate


(iv) 0016 +LDA NUM2

• Loads the contents of address of NUM2 into the accumulator.

• Address = 00100


000000 1 1 0 0 0 1 00100

Opcode = 03100100

• Addressing Mode Direct


3. Explain what is meant by Program Relocation and use an example to show how

it is done.

Ability of loading a program anywhere in memory without stating the actual starting

address.

It is done by writing modification records e.g. for an instruction

006 +JSUB RDREC 4B101036 the modification record would be

M00000705

4. (a) How are literals different from immediate operands?

With immediate addressing the operand value is assembled as part of the

machine instruction. With a literal the assembler generates the specified value

as a constant at some other memory location. The address of this generated

constant is used as the target address for the machine instruction.

(b) Use examples to explain the difference between Absolute expressions and

Relative expressions.

An absolute expression contains only absolute terms or relative terms so long

as the relative terms occur in pairs and the terms in each pair have opposite

signs. e.g. MAXLEN EQU BUFEND – BUFFER

A relative expression is one in which all the relative terms except one can be

paired. The remaining unpaired relative term must have a positive sign e.g.

MAXLEN EQU BUFEND + 100

5 (a) Modify the SIC piece of code below by declaring the variables first so that

it can be assembled in one pass. 1000 COPY START 1000 1000 EOF BYTE C’EOF’ 454F46 1003 THREE WORD 3 000003 1006 ZERO WORD 0 000000 1009 RETADR RESW 1 100C LENGTH RESW 1 100F BUFFER RESB 4096 200F FIRST STL RETADR 141009 2012 CLOOP JSUB RDREC 480000 2015 LDA LENGTH 00100C 2018 COMP ZERO 281006 201B JEQ ENDFIL 300000 201E JSUB WRREC 480000 2021 J CLOOP 3C2012 2024 ENDFIL LDA EOF 001000 2027 STA BUFFER 0C100F 202A LDA THREE 001003 202D STA LENGTH 0C100C 2030 JSUB WRREC 482061 2033 LDL RETADR 081009 2036 RSUB 4C0000

(b) Generate the text listing for the modified code assuming a one pass

assembler that generates object code on an output device.

H^COPY ^00100^001039

T^001000^09^454F46^000003^000000

T^00200F^15^141009^480000^00100C^281006^300000^480000^3C2012

T^00201C^02^2024

T^002024^15^001000^0C100F^001003^0C100C^480000^081009^4C0000

E^00200F

MAKERERE UNIVERSITY







DATE: 25th January 2007 TIME: 4.00p.m.-7.00 p.m.

EXAMINATION INSTRUCTIONS

1. SECTION A IS COMPULSORY (40 MARKS)

2. ATTEMPT (03) QUESTIONS FROM SECTION B. ALL QUESTIONS CARRY

EQUAL MARKS (60 MARKS)

3. DO NOT OPEN THIS EXAM UNTIL YOU ARE TOLD TO DO SO

4. ALL ROUGH WORK SHOULD BE IN YOUR ANSWER BOOKLET.


1. (a) Explain briefly what you understand by Systems Programming. (2 marks)

(b) Give and briefly explain the uses of any three Systems Software. (3 marks)

2. (a) Briefly explain the uses of the following computer components:

(i) CPU (ii) Memory (iii) Memory Interfaces (4 marks)

3. Use examples to explain the difference between Direct and Indirect addressing of the

SIC/XE machine. (4 marks)

4. Explain any four directives used by the SIC/XE machine. (4 marks)

5. (a) What do you understand by Assembling a program? (1 mark)

(b) A program in Assembler language is usually assembled using two passes.

Briefly explain what is done in each pass. (5 marks)


(i) Program Relocation (ii) A Literal Pool

(iii) A Relocation bit (iv) A bit mask (4 marks)

7. Use an example to explain what is meant by Programs with Overlay structures.

(2 marks)

8. Explain briefly the difference between a Program Block and a Control Section.

(4 marks)

9. Briefly explain the following:

(i) PROGADDR (ii) ESTAB

(iii) CSADDR iv) SEGTAB (4 marks)

10. Explain the following and give examples where possible.

(i) A Grammar (ii) Lexical Analysis (iii) Syntactic Analysis.

(3 marks)


11. (a) Write a simple SIC/XE program that gets two integers NUM1 and NUM2

from memory, compares the two integers and adds them up if NUM1 is

greater than NUM2 or subtracts NUM1 from NUM2 if NUM1 is less than

NUM2. Store the answer in memory at a memory location RESULT.

(10 marks)

(b) Translate the first three instructions of the above program into machine

language and generate the machine listing for the written code. (10 marks)

12. Consider the code below. � � � � � � � � �

� � � � � � � � � � � � � �

� � � � � � ! " � #

� � � � � � ! " � #

� � � � � � $ % � & � � � � � �

� � � � ! & '

� � � ' �

� % � ( � ! � � � �

� � $ % � & � ) � � � �

� * % � � � � �

� � � ! � � � � � � � + � , � � � ,

� � � � � � & � � � �

� � � � �

� � � � � � � ! " � #

� � $ % � & � ) � � � �

� � � � � � "

� � - + � , � � � ,

� � � � � � � � � � ) �

� � � � ! " � # � � � ) �

� � � & � � � � � � � � � . �

� � � � & � � ! � � ( & -

� ' � / � � ! � ( & � & � � ! � 0 � & � � � �

(a) Write machine code for the instructions at addresses 0000, 0003, and 000D.

(6 marks)

(b) What does the Directive at address 1036 mean? (2 marks)

(c) Write all the Modification Records that may be needed in this piece of code.

Why don’t the other remaining instructions need Modification Records?

(8 marks)

(d) Assume that RDREC and WRREC are in different control sections and that

the variables LENGTH, BUFFER and BUFEND are referenced by these other

control sections, insert some extra lines of code in the above program to reflect

this assumption. (4 marks)

13. The program below consists of two different control sections that are to be linked

together.

(a) Write down the machine code for the instructions and directives for control

section PROGA. (5 Marks)

(d) Write down the object code (Text Listing) for PROGA as it would appear on

an output device. (5 Marks)

(e) Draw an external symbol table for this program assuming that PROGA is to be

loaded in memory from a memory location 2500 and that PROGB will

immediately follows PROGA (4 Marks)

(d) Using the assumptions made in part (c) above calculate the values for REF6

and REF7. (6 Marks)

0000 PROGA START 0 EXTDEF LISTA,ENDA EXTREF LISTB, ENDB . 0020 REF1 LDA LISTA 0023 REF2 +LDT LISTB + 4 0040 LISTA EQU * 005A REF6 WORD LISTA + LISTB 005D REF7 WORD ENDB-LISTB+ LISTA END

0000 PROGB START 0 EXTDEF LISTB,ENDB EXTREF LISTA, ENDA 0036 REF1 +LDA LISTA 003A REF2 LDT LISTB + 4 0060 LISTB EQU * 0070 ENDB EQU * 0076 REF6 WORD LISTA + LISTB 0079 REF7 WORD ENDB-LISTB+ LISTA END

14. A diagram of Overlay control sections is as shown below and the number of bytes

occupied by each control section in memory is also given.

1

2 7 4

3 8 5 6

Control Length Control

Length

Section (bytes) Section (bytes)

A 3000 F 1900

B 1800 G 400

C 4700 H 800

D 2900 X 1000

E 800 Y 800

(a) Write a series of statements that can be used to describe the above structure.

(5 Marks)

(b) Draw a segment table that will describe this structure assuming that control

section A is to be loaded from address 5000. (10 Marks)

(c) Explain the following:

(i) Overlay Manager (ii) Segment File (3 Marks)

(d) What other control sections will be needed in memory in order to run Program

Y? (2 Marks)

15. (a) Assume the PASCAL grammar and draw a top down parse tree for this

expression:

FOR I := 1 to 10 DO SUM := SUM + I (10 Marks)

A

B

C/D

E/F

G H

X

Y

(c) Use a bottom up method to parse the following statement using the operator

precedence method.

ANS := ALPHA DIV (BETA + GAMMA) – DELTA; (10 marks)


MNEMONIC FORMAT OPCODE EFFECT Notes ADD m 3/4 18 A (A) + (m...m+2) ADDF m 3/4 58 F (F) + (m...m+5) X F ADDR r1,r2 2 90 r2 (r2) + (r1) X AND m 3/4 40 A (A) & (m…m+2) CLEAR r1 2 B4 r1 0 X COMP m 3/4 28 (A) : (m…m+2) C COMPF m 3/4 88 (F) : (m…m+5) X F COMPR r1,r2 2 A0 (r1) : (r2) X DIV m 3/4 24 A (A) / (m….m+2) DIVF m 3/4 64 F (F) / (m….m+5) X F DIVR r1,r2 2 9C r2 (r2) / (r1) X FIX 1 C4 A (F) convert to X F FLOAT 1 C0 F (A) convert to X F HIO 1 F4 Halt I/O channel number P X J m 3/4 3C PC m JEQ m 3/4 30 PC m if CC set to = JGT m 3/4 34 PC m if CC set to > JLT m 3/4 38 PC m if CC set to < JSUB m 3/4 48 L (PC); PC m LDA m 3/4 00 A (m..m+2) LDB m 3/4 68 B (m..m+2) X LDCH m 3/4 50 A [rightmost byte] LDF m 3/4 70 F (m..m+5) X F LDL m 3/4 08 L (m..m+2) LDS m 3/4 6C S (m..m+2) X LDT m 3/4 74 T (m..m+2) X LDX m 3/4 04 X (m..m+2) LPS m 3/4 D0 Load processor status from P X MUL m 3/4 20 A (A) * (m..m+2) MULF m 3/4 60 F (A) * (m..m+5) X F MULR r1,r2 2 98 r2 (r2) * (r1) X NORM 1 C8 F (F) normalised X F OR m 3/4 44 A (A) | (m..m+2) RD m 3/4 D8 A [rightmost byte] P RMO r1,r2 2 AC r2 (r1) X RSUB 3/4 4C PC (L) SHIFTL r1,n 2 A4 r1 (r1) left shift n X SHIFTR r1,n 2 A8 r1 (r1) right shift n X SIO 1 F0 Start I/O channel number P X SSK m 3/4 EC Protection key for address P X STA m 3/4 0C m..m+2 (A) STB m 3/4 78 m..m+2 (B) X STCH m 3/4 54 m (A) [rightmost STF m 3/4 80 m..m+5 (F) X F STI m 3/4 D4 Interval timer value P X STL m 3/4 14 m..m+2 (L) STS m 3/4 7C m..m+2 (S) X STSW m 3/4 E8 m..m+2 (SW) P STT m 3/4 84 m..m+2 (T) X STX m 3/4 10 m..m+2 (X) SUB m 3/4 1C A (A) - (m..m+2) SUBF m 3/4 5C F (F) - (m..m+5) X F SUBR r1,r2 2 94 r2 (r2) – (r1) X SVC n 2 B0 Generate SVC Interrupt X TD m 3/4 E0 Test Device specified by P C TIO 1 F8 Test I/O channel number P X TIX m 3/4 2C X (X) + 1; (X) : C

TIXR r1 2 B8 X (X) + 1; (X) : (r1) X C WD m 3/4 DC Device specified by (m)

(A) [rightmost byte]

P

OPERATOR PRECEDENCE TABLE



; > > < < < > < < <

: > < >

, =

:= > = > < < < < < < <

+ > > > > > > < < < > < <

- > > > > > > < < < > < <

* > > > > > > > > < > < <

DIV > > > > > > > > < > < <

( > < < < < < < = < <

) > > > > > > > > >

Id > > > > > > > = > > > > >

Int > > > > > > > > >

THE PASCAL GRAMMAR

18. <prog> ::= PROGRAM <prog-name> VAR <dec-list> BEGIN <stmt-list> END








26. <assign> ::= id := <exp>



29. <factor> ::= id | int | ( <exp>)






MAKERERE UNIVERSITY







DATE: 25th January 2007 TIME: 4.00p.m.-7.00 p.m.








1. (a) Explain briefly what you understand by Systems Programming. (2 marks)

(b) Give and briefly explain the uses of any three Systems Software. (3 marks)

2. (a) Briefly explain the uses of the following computer components:

(i) CPU (ii) Memory (iii) Memory Interfaces (4 marks)

3. Use examples to explain the difference between Direct and Indirect addressing of the

SIC/XE machine. (4 marks)

4. Explain any four directives used by the SIC/XE machine. (4 marks)

5. (a) What do you understand by Assembling a program? (1 mark)

(b) A program in Assembler language is usually assembled using two passes.

Briefly explain what is done in each pass. (5 marks)


(i) Program Relocation (ii) A Literal Pool

(iii) A Relocation bit (iv) A bit mask (4 marks)

7. Use an example to explain what is meant by Programs with Overlay structures.

(2 marks)

8. Explain briefly the difference between a Program Block and a Control Section.

(4 marks)

10. Briefly explain the following:

(i) PROGADDR (ii) ESTAB

(iii) CSADDR iv) SEGTAB (4 marks)

11. Explain the following and give examples where possible.

(i) A Grammar (ii) Lexical Analysis (iii) Syntactic Analysis.

(3 marks)


11. (a) Write a simple SIC/XE program that gets two integers NUM1 and NUM2

from memory, compares the two integers and adds them up if NUM1 is

greater than NUM2 or subtracts NUM1 from NUM2 if NUM1 is less than

NUM2. Store the answer in memory at a memory location RESULT.

(10 marks)

(b) Translate the first three instructions of the above program into machine

language and generate the machine listing for the written code. (10 marks)

12. Consider the code below. � � � � � � � � �

� � � � � � � � � � � � � �

� � � � � � ! " � #

� � � � � � ! " � #

� � � � � � $ % � & � � � � � �

� � � � ! & '

� � � ' �

� % � ( � ! � � � �

� � $ % � & � ) � � � �

� * % � � � � �

� � � ! � � � � � � � + � , � � � ,

� � � � � � & � � � �

� � � � �

� � � � � � � ! " � #

� � $ % � & � ) � � � �

� � � � � � "

� � - + � , � � � ,

� � � � � � � � � � ) �

� � � � ! " � # � � � ) �

� � � & � � � � � � � � � . �

� � � � & � � ! � � ( & -

� ' � / � � ! � ( & � & � � ! � 0 � & � � � �

(a) Write machine code for the instructions at addresses 0000, 0003, and 000D.

(6 marks)

(b) What does the Directive at address 1036 mean? (2 marks)

(c) Write all the Modification Records that may be needed in this piece of code.

Why don’t the other remaining instructions need Modification Records?

(8 marks)

(d) Assume that RDREC and WRREC are in different control sections and that

the variables LENGTH, BUFFER and BUFEND are referenced by these other

control sections, insert some extra lines of code in the above program to reflect

this assumption. (4 marks)

16. The program below consists of two different control sections that are to be linked

together.

(a) Write down the machine code for the instructions and directives for control

section PROGA. (5 Marks)

(f) Write down the object code (Text Listing) for PROGA as it would appear on

an output device. (5 Marks)

(g) Draw an external symbol table for this program assuming that PROGA is to be

loaded in memory from a memory location 2500 and that PROGB will

immediately follows PROGA (4 Marks)

(d) Using the assumptions made in part (c) above calculate the values for REF6

and REF7. (6 Marks)

0000 PROGA START 0 EXTDEF LISTA,ENDA EXTREF LISTB, ENDB . 0020 REF1 LDA LISTA 0023 REF2 +LDT LISTB + 4 0040 LISTA EQU * 005A REF6 WORD LISTA + LISTB 005D REF7 WORD ENDB-LISTB+ LISTA END

0000 PROGB START 0 EXTDEF LISTB,ENDB EXTREF LISTA, ENDA 0036 REF1 +LDA LISTA 003A REF2 LDT LISTB + 4 0060 LISTB EQU * 0070 ENDB EQU * 0076 REF6 WORD LISTA + LISTB 0079 REF7 WORD ENDB-LISTB+ LISTA END

17. A diagram of Overlay control sections is as shown below and the number of bytes

occupied by each control section in memory is also given.

1

2 7 4

3 8 5 6

Control Length Control

Length

Section (bytes) Section (bytes)

A 3000 F 1900

B 1800 G 400

C 4700 H 800

D 2900 X 1000

E 800 Y 800

(a) Write a series of statements that can be used to describe the above structure.

(5 Marks)

(b) Draw a segment table that will describe this structure assuming that control

section A is to be loaded from address 5000. (10 Marks)

(c) Explain the following:

(i) Overlay Manager (ii) Segment File (3 Marks)

(d) What other control sections will be needed in memory in order to run Program

Y? (2 Marks)

18. (a) Assume the PASCAL grammar and draw a top down parse tree for this

expression:

FOR I := 1 to 10 DO SUM := SUM + I (10 Marks)

A

B

C/D

E/F

G H

X

Y

(d) Use a bottom up method to parse the following statement using the operator

precedence method.

ANS := ALPHA DIV (BETA + GAMMA) – DELTA; (10 marks)


MNEMONIC FORMAT OPCODE EFFECT Notes ADD m 3/4 18 A (A) + (m...m+2) ADDF m 3/4 58 F (F) + (m...m+5) X F ADDR r1,r2 2 90 r2 (r2) + (r1) X AND m 3/4 40 A (A) & (m…m+2) CLEAR r1 2 B4 r1 0 X COMP m 3/4 28 (A) : (m…m+2) C COMPF m 3/4 88 (F) : (m…m+5) X F COMPR r1,r2 2 A0 (r1) : (r2) X DIV m 3/4 24 A (A) / (m….m+2) DIVF m 3/4 64 F (F) / (m….m+5) X F DIVR r1,r2 2 9C r2 (r2) / (r1) X FIX 1 C4 A (F) convert to X F FLOAT 1 C0 F (A) convert to X F HIO 1 F4 Halt I/O channel number P X J m 3/4 3C PC m JEQ m 3/4 30 PC m if CC set to = JGT m 3/4 34 PC m if CC set to > JLT m 3/4 38 PC m if CC set to < JSUB m 3/4 48 L (PC); PC m LDA m 3/4 00 A (m..m+2) LDB m 3/4 68 B (m..m+2) X LDCH m 3/4 50 A [rightmost byte] LDF m 3/4 70 F (m..m+5) X F LDL m 3/4 08 L (m..m+2) LDS m 3/4 6C S (m..m+2) X LDT m 3/4 74 T (m..m+2) X LDX m 3/4 04 X (m..m+2) LPS m 3/4 D0 Load processor status from P X MUL m 3/4 20 A (A) * (m..m+2) MULF m 3/4 60 F (A) * (m..m+5) X F MULR r1,r2 2 98 r2 (r2) * (r1) X NORM 1 C8 F (F) normalised X F OR m 3/4 44 A (A) | (m..m+2) RD m 3/4 D8 A [rightmost byte] P RMO r1,r2 2 AC r2 (r1) X RSUB 3/4 4C PC (L) SHIFTL r1,n 2 A4 r1 (r1) left shift n X SHIFTR r1,n 2 A8 r1 (r1) right shift n X SIO 1 F0 Start I/O channel number P X SSK m 3/4 EC Protection key for address P X STA m 3/4 0C m..m+2 (A) STB m 3/4 78 m..m+2 (B) X STCH m 3/4 54 m (A) [rightmost STF m 3/4 80 m..m+5 (F) X F STI m 3/4 D4 Interval timer value P X STL m 3/4 14 m..m+2 (L) STS m 3/4 7C m..m+2 (S) X STSW m 3/4 E8 m..m+2 (SW) P STT m 3/4 84 m..m+2 (T) X STX m 3/4 10 m..m+2 (X) SUB m 3/4 1C A (A) - (m..m+2) SUBF m 3/4 5C F (F) - (m..m+5) X F SUBR r1,r2 2 94 r2 (r2) – (r1) X SVC n 2 B0 Generate SVC Interrupt X TD m 3/4 E0 Test Device specified by P C TIO 1 F8 Test I/O channel number P X TIX m 3/4 2C X (X) + 1; (X) : C

TIXR r1 2 B8 X (X) + 1; (X) : (r1) X C WD m 3/4 DC Device specified by (m)

(A) [rightmost byte]

P

OPERATOR PRECEDENCE TABLE



; > > < < < > < < <

: > < >

, =

:= > = > < < < < < < <

+ > > > > > > < < < > < <

- > > > > > > < < < > < <

* > > > > > > > > < > < <

DIV > > > > > > > > < > < <

( > < < < < < < = < <

) > > > > > > > > >

Id > > > > > > > = > > > > >

Int > > > > > > > > >

THE PASCAL GRAMMAR

35. <prog> ::= PROGRAM <prog-name> VAR <dec-list> BEGIN <stmt-list> END








43. <assign> ::= id := <exp>



46. <factor> ::= id | int | ( <exp>)






MAKERERE UNIVERSITY







DATE: 4th December 2007 TIME: 4.00p.m.-7.00 p.m.








1. (a) Give and briefly explain the uses of any three Systems Software.

• Compilers translate high level programs into machine code.

• Assemblers translate Assembler Language into machine code.

• Loaders they prepare machine language programs for execution

• Macro processors allow programmers to use abbreviations

• Operating Systems and file systems allow flexible storing and retrieval of

information. (3 marks)

(b) How does Systems Software differ from Application Software?

Application programs are written by the user for the purpose of solving

particular User problems using a computer as a tool whereas the Systems

Software is to make more effective use of the computer. (1 mark)

2. Explain briefly:

(i) The Format of a machine language instruction.

It is the arrangement of an instruction with respect to assigning meaning to its

various groups of bits. It consists of the Operation Code which is a portion

that specifies what the instruction does and the Operand which is any piece of

data or address that is required by the instruction to complete its execution.

(2 marks)

(ii) The Format of an Assembler Language instruction.

Label: Mnemonic Operand, Operand, ; remarks

Label: It is a symbol assigned to the address of the first byte of the instruction

in which it appears.

A mnemonic is the abbreviation for the real instruction.

The Operand(s) is the data or the address needed to complete the instruction.

Remarks are for documenting the program. (2

marks)

3. Use examples to help you explain any 3 addressing modes for the Intel 8085

microprocessor.

(Any 3 of the following)

• Immediate Addressing: Information is part of the instruction

• Direct addressing: The address is part of the instruction.

• Register addressing: The operand is in the register and the register’s address is part

of the instruction.

• Indirect Addressing: The address is in the location whose address is specified as part

of the instruction.

• Base addressing: The address is formed by adding the contents of a memory location

or register to a number called a displacement which is part of the instruction.

• Indexing: It is a process of incrementing or decrementing an address as the computer

sequences through a set of consecutive or evenly spaced addresses. (6 marks)

4. Consider the following INTEL 8085 Assembler Code.

0045 NUM: DB 10, 20, 30 3020 MAK: DB 50, 60

(a) Explain what each of the following instruction does and give the contents

of the registers involved in each case.

(i) LDA NUM

Loads contents of the address NUM into the accumulator; A = 10

(2 marks)

(ii) LHLD NUM+1

Loads contents of NUM+1 and NUM+2 into the register pair HL

H = 30 L = 20 (2 marks)

(iii) LXI D, NUM + 1

Loads register pair DE with the address of NUM + 1

D = 00 E = 46 (2 marks)

(b) Assume that each of the instructions in part (a) above were executed and

explain what each of the following instruction does and the contents of the

registers involved in each case.

(i) INX D

Increment register pair DE; i.e. D = 00; E = 47 (2 marks)

(ii) INR M

Increment contents of a memory location whose address is in the

register pair H.L i.e. M = 51 (2 marks)

(iii) LDAX D

Load the Accumulator with contents of a memory location whose

address is in the DE register pair; A = 30 (2 marks)

13. Give the Mnemonics of any five registers found in the SIX/XE and state their

uses.

(Any 5 of the following) A Accumulator; used for arithmetic Operations

X Index register; used for addressing

L Linkage register; the Jump to subroutine (JSUB) instruction stores the return

address in this register.

PC Program Counter; Contains the address of the next instruction to be fetched

for execution.

SW Status word; contains the condition codes

B Base register, used for addressing

S General Working register, no special use.

T General Working register, no special use.

F Floating Point Accumulator. (5 marks)

6. Assume the following code for the SIC machine. The Operation Code for the

Mnemonic STL is 14

0022 STL Buffer, X 2020 Buffer: RESW 1

Derive machine code for the instruction at address 0022.

6 1 1 1 1 1 1 20

0001 01 1 1 1 0 0 1 02020 = 17B02020 (2 marks)


7. (i) What is program relocation?

Ability of a program to be able to run successfully in memory without

considering the address where it is going to be loaded. (2 marks)

(ii) Explain why some instruction need modification records during program

relocation and others do not.

Instruction that have been assembled using addresses are the ones that need to

be modified; format 1, format 2 do need have address operands; and format 3

instructions that have been assembled using program counter relative or base

relative are similarly not affected because the displacements do not change

when the program is loaded. (2 marks)


(i) PROGADDR

The beginning address in memory where the linked program is loaded.

(1 marks)

(ii) ESTAB

External symbol used to store names and addresses of each external symbol in

the set of control sections being loaded. It also indicates in which control

section the symbol is defined. (1 marks)

(iii) CSADDR

The starting address assigned to the control section currently being scanned

by the loader. (1 marks)


9. (a) Briefly explain the uses of the components of the of the Control Unit of a

typical microprocessor.

• The Program Counter (PC) :It holds the address of the main memory

location from which the next instruction is to be fetched.

• Instruction Register (IR) Receives the instruction when it is brought from

memory and holds it while it gets decoded and executed.

• Processor Status Word (PSW) contains condition flags which indicate the

current status of the CPU and the important characteristics of the result of

the previous instruction.

• Stack Pointer (SP): Accesses a special part of memory called a stack. It is

used to temporarily store important information while sub routines are

being executed. It hold the address at the top of the stack. (8 marks)

(b) (i) What is an interface

Circuitry needed to connect the bus to a device. (1 mark)

(ii) Give the functions of the I/O interface

• Buffer data onto/off the system bus

• Receive commands from the CPU

• Transmit information from their devices to the CPU (3 marks)

(iii) Give the functions of the memory interface.

• Needed to decode the address of the memory location being

accessed.

• Buffer data onto/off the bus.

• Contain circuitry to perform memory reads or write (3 marks)

(c) Give and explain the components of the Processor Status Word for the

Intel 8085 microprocessor.

The PSW for the Intel 8085 has 5 flags.

• Zero (Z): set when the result of the operation is zero.

• Sign (S): set when sign of the result is negative.

• Parity (P): When the parity of the bits in the result is even.

• Carry (C): Addition resulted into a carry or subtraction or comparison

resulted into a borrow.

• Auxiliary Carry (AC) Carry in BCD arithmetic. (5 marks)

10. (a) An array MYARRY contains 10 integers (negative and positive). Write an

Intel 8085 Assembler language program that will add up all the integers

of this array and store the answer at a memory location GREATER if the

answer is greater than 100 or at a memory location LESS if the answer is

less than 100.

MVI D, 10

LXI H, MYARRAY

S Z AC P C

MOV A, M

LOOP: INX H

ADD M

DCR D

JNZ LOOP

CPI 100

JP MORE

STA LESS

JMP OUT

MORE: STA GREATER

MYARRAY DS 10

GREATER: DS 1

LESS DS 1

(10 Marks)

(b) Write an equivalent program for the same problem using the SIC/XE

machine code.

LDS #100

LDT # 10

LDA #0

LDX #1

AGAIN: ADD MYARRAY, X

TIXR T

JLT AGAIN

COMPR A, S

JLT SMALL

STA GREATER

J OUT

SMALL: STA LESS

OUT:

MYARRAY: RESW 10

GREATER: RESW 1

LESS: RESW 1 (10 Marks)

11. Given below is a SIC/XE program PROG.

LDS # 10

LDA #0

LDX #1

AGAIN: ADD ALPHA, X

TIXR S

JLT AGAIN

STA RESULT

ALPHA: RESW 10

RESULT: RESW 1

(iv) Study the program carefully and explain briefly what is being

accomplished.

The program adds up 10 integers in an array ALPHA and stores the answer at

a memory location RESULT (3 Marks)

(v) Assuming that it is to be loaded in memory from 0058H, generate the

corresponding addresses for each instruction and write the corresponding

machine code using the operation code table below.

Mnemonic Opcode

LDS 6C

LDA 00

LDX 04

ADD 18

TIXR B8

STA 0C

JLT 38

S 4

0058 LDS #10 6D000A

005B LDA #0 010000

005E LDX #1 050001

0061 AGAIN: ADD ALPHA,X 1BA008

0064 TIXR S B840

0066 JLT AGAIN 3B2FF8

0069 STA RESULT 0F201E

006C ALPHA RESW 10

008A RESULT RESW 1

(17 Marks)

12. (a) Two passes are usually used to translate an Assembler Language program to machine code; explain briefly what is accomplished in each

pass and state the tables and variables used in each pass.

• Pass 1 (Define symbols)

• Scans the source program for label definitions using the LOCATION

COUNTER and assigns addresses to all statements in the program.

• Saves the addresses assigned to all labels for use in pass 2 in the

SYMBOL TABLE

• Perform some processing of the assembler directives.

• Pass 2 (assemble the instructions and generate machine code)

• Assembles the instruction using the Symbol Table (SYMTAB) and the

operations Code Table (OPTAB).

• Generate data values defined by BYTE, WORD, etc.

• Perform processing of the directives not done during pass 1.

• Write the object program and assembly listing onto some output device

which will later be loaded in memory for execution. (5 marks)

(c) Consider the code given below.

0000 EXAM START 0

0000 FIRST STL RETAD 17202D

0003 LDB #LENG 69202D

BASE LENGT

0006 CLOOP +JSUB RDREC 4B10103

000A LDA LENGT 032026

000D COMP #0 290000

0010 JEQ ENDFIL 332007

0013 +JSUB WRREC 4B10105

0017 J CLOOP 3F2FEC

001A ENDFIL LDA EOF 032010

001D STA BUFFE 0F2016

0020 LDA #3 010003

0023 STA LENGT 0F200D

0026 +JSUB WRREC 4B10105

002A J @RETA 3E2003

002D EOF BYTE C’EOF’ 454F46

0030 RETAD RESW 1

0033 LENGT RESW 1

0036 BUFFE RESB 4096

(i) Some instructions in the code above will change when the program

is loaded in memory from a different memory location other than

0000. Give the addresses of the instructions which will change and

explain why the others will not change.

Instructions at addresses 0006, 0013 and 0026 will change. The others

will not change because their machine codes do not contain addresses.

(3 marks)

(i) Generate the object code for the program above. H^ EXAM ^000000^ 001036

T^000000^1D^17202D^69202D^4B101036^032026^290000^332007^4B10105D^3F2FEC^032010

T^000001D^13^0F2016^010003^0F200D^4B10105D^3E2003^454F46

M^000007^05

M^000014^05

M^000027^05

E^000000 (12 marks)

13. Consider the program code below written in the simple SIC format.

(a) Modify the program so that it can be assembled in one pass. (10 marks)

1000 EXAM START 1000

1000 INDEX RESW 1

1003 ALPHA RESW 4

100F BETA RESW 4

101B GAMMA RESW 4

1027 THREE WORD 3 000003

102A ZERO WORD 0 000000

102D K300 WORD 12 00012C

1030 LDA ZERO 00102A

1033 STA INDEX 0C1000

1036 LOOP LDX INDEX 041000

1039 LDA ALPHA,X 009003

103C ADD BETA,X 18900F

103F STA GAMMA,X 0C901B

1042 LDA INDEX 001000

1045 ADD THREE 181027


104B COMP K300 28102D

104E JLT LOOP 381036

(10 marks)

(b) Use the Load and Go method to show how the above program will appear

in memory.

1000 1001 1002 1003 1004 1005 1006 1007 1008 1009 100A 100B 100C 100D 100E 100F

1010 1011 1012 1013 1014 1015 1016 1017 1018 1019 101A 101B 101C 101D 101E 101F

00 00 03 00 00 00 00 01 2C

1020 1021 1022 1023 1024 1025 1026 1027 1028 1029 102A 102B 102C 102D 102E 102F

00 10 2A 0C 10 00 04 10 00 00 90 03 18 90 0F 0C

1030 1031 1032 1033 1034 1035 1036 1037 1038 1039 103A 103B 103C 103D 103E 103F

90 1B 00 10 00 18 10 27 0C 10 00 28 10 2D 38 10

1040 1041 1042 1043 1044 1045 1046 1047 1048 1049 104A 104B 104C 104D 104E 104F

36

1050 1051 1052 1053 1054 1055 1056 1057 1058 1059 105A 105B 105C 105D 105E 105F

(6 marks)

(c) Explain the following

(i) Programs with Overlay Structures.

They are programs that are designed to execute in such a way that if

both or all of them are not needed in memory at the same time, one

can execute first and the other will execute in the same memory space

after the first one has been executed. (2 marks)

(ii) The meaning of the expression: MYNUM EQU * The current value of the location counter is assigned to the variable

MYNUM (2 marks)

14. (a) (i) Briefly explain the difference between Program Blocks and

Control Sections.

Control sections differ from program blocks in that they are handled

separately by the assembler. It is not necessary for all control sections

in a program to be assembled at the same time. Each control section is

independent of another whereas Program Blocks are sections of code

re-arranged in the same unit. (2 marks)

(ii) What are the directives used to indicate a new Program Block and

a new Control section respectively in a program?

The directive USE indicates a new Program Block while the Directive

CSECT announces a new Control Section. (2 marks)

(b) Assume that the following piece of code is in a program block called

PROGA.

001C LDT NUM

002C Number: RESB 1

(i) If the beginning address of PROGA is 000A and the code at

address 002C is the last directive in PROG A, what is the length of

Program Block PROGA?

002D – 000A = 0023 (2 marks)

(ii) Assume that NUM is defined at address 000E in the next Program

Block PROGB which begins at address 0000H. If the Operation

Code for the instruction LDT is 74, derive the machine instruction

for the Assembler instruction at address 001C in PROGA.

The displacement = 002D – 001F + 000E = 01C

6 1 1 1 1 1 1 12

0111 0111 0010 01C = 77201C (4 marks)

(c) Explain any 5 Loader Options that can be used to modify the standard

processing of a program.

• INCLUDE: Allows the selection of alternative sources of input.

• DELETE Allows the user to delete external symbols or entire control

section.

• CHANGE: To change external symbols

• LIBRARY: Specifies alternative libraries to be searched.

• NOCALL: To exclude some functions that come with the library search

• Options to control outputs from the loader. The user can specify whether

an output is needed or not.


• Options to specify whether external references will be resolved by

library search.

• An option to specify the location at which execution is to begin thus

overriding any information given in the object program.

• An option to control whether or not the loader should attempt to execute

the program if errors are detected during the load. (10 marks)

15. Given below are two control sections PROG1 and PROG2 both loaded in

memory.

(a) Give the operands for the Directives EXTREF and EXTDEF in each

control section.

Prog1 Prog2

EXDEF NUM1 EXDEF STORE2

EXTREF STORE2 EXTREF NUM1 (2 marks)

(b) Derive machine code for the instructions at addresses 0020, 0027, 0060 in

PROG1 and 0036, 003D and 007C in PROG2 assuming that the operation

codes for the instructions LDT and STT are 74 and 84 respectively.

AT 0020: The displacement = 0040 – 23 = 01D

6 1 1 1 1 1 1 12

0111 0111 0010 01D = 77201D (1 mark)

AT 0027: The target address is not known; we use format 4

6 1 1 1 1 1 1 20

1000 0111 0001 00000 = 87100000 (1 mark)

AT 0060: 0000 – 0040 = -000040 = FFFFC0 (1 mark)

AT 0036: NUM is not known, format 4 is used

0111 0111 0001 00000 = 77100000 (1 mark)

AT 003D Displacement = 0060 – 0040 = 020

1000 1011 0010 020 = 872020 (1 mark)

AT 007C: 000060 – 000000 = 000060 (1 mark)

(c) Write the object code for the two Control Sections.

H^PROG1 ^000000^000063



D^NUM1 ^000040

R^STORE2

T^000020^0B^77201D ^8710000

T^000060^03^FFFFC0

M^000028^05^+STORE2

M^000060^06^+STORE2

M^000060^06^-PROG1

E^000020 (4 marks)

H^PROG2 ^000000^00007F

D^STORE2 ^000060

R^NUM1

T^000036^0A^77100000^872020

T^00007C^03^000060

M^000037^05^+NUM1

M^00007C^06^+PROG2

M^00007C^06^-NUM1

E (4 marks)

(d) Assume that the first control section PROG1 is loaded in memory at

address 1200 and that PROG2 is loaded immediately after PROG1,

calculate the value of RESULT

( 001200 + 63 + 60) – (001200 + 40) = 000083 (4 marks)

0000

PROG1 START 0

EXTDEF

EXTREF

.

0020 LDT NUM1 77201D

0027 +STT STORE2 8710000

0

0040 NUM1 RESW 1

0060 RESULT WORD STORE2 –

NUM1

FFFFC0

END

0000 PROG2 START 0

EXTDEF

EXTREF

.

0036 +LDT NUM1 77100000

003D STT STORE2 872020

0060 STORE2 RESW 1

007C RESULT WORD STORE2 -

NUM1

000060

END

MAKERERE UNIVERSITY







DATE: 4th December 2007 TIME: 4.00p.m.-7.00 p.m.








1. (a) Give and briefly explain the uses of any three Systems Software. (3 marks)

(b) How does Systems Software differ from Application Software? (1 mark)

2. Explain briefly:

(i) The Format of a Machine Language instruction. (2 marks)

(ii) The Format of an Assembler Language instruction. (2 marks)

3. Use examples to help you explain any 3 addressing modes for the Intel 8085

microprocessor. (6 marks)

4. Consider the following INTEL 8085 Assembler Code.

0045 NUM: DB 10, 20, 30 3020 MAK: DB 50, 60

(a) Explain what each of the following instruction does and give the contents of

the registers involved in each case.

(i) LDA NUM (ii) LHLD NUM+1 (iii) LXI D, NUM + 1 (6 marks)

(b) Assume that each of the instructions in part (a) above were executed and

explain what each of the following instructions does and the contents of the

registers involved in each case.

(i) INX D (ii) INR M (iii) LDAX D (6 marks)

5. Give the Mnemonics (abbreviations) of any five registers found in the SIX/XE and

state their uses. (5 marks)

6. Assume the following code for the SIC machine. The Operation Code for the

Mnemonic STL is 14

0022 STL Buffer, X 2020 Buffer RESW 1

Derive machine code for the instruction at address 0022. (2 marks)

7. (i) What is program relocation? (2 marks)

(ii) Explain why some instruction need modification records during program

relocation and others do not. (2 marks)


(i) PROGADDR (ii) ESTAB (iii) CSADDR (3 marks)


9. (a) Briefly explain the uses of the components of the of the Control Unit of a

typical microprocessor. (8 marks)

(b) (i) What is an interface (1 mark)

(ii) Give the functions of the I/O interface (3 marks)

(iii) Give the functions of the memory interface. (3 marks)

(c) Give and explain the components of the Processor Status Word for the Intel

8085 microprocessor. (5 marks)

10. (a) An array MYARRY contains 10 integers (negative and positive). Write an

Intel 8085 Assembler language program that will add up all the integers of this

array and store the answer at a memory location GREATER if the answer is

greater than 100 or at a memory location LESS if the answer is less than 100.

(10 Marks)

(b) Write an equivalent program using the SIC/XE machine code for the same

problem. (10 Marks)

11. Given below is a SIC/XE program PROG.

LDS # 10 LDA #0 LDX #1 AGAIN: ADD ALPHA, X TIXR S JLT AGAIN STA RESULT ALPHA: RESW 10 RESULT: RESW 1

(vi) Study the program carefully and explain briefly what is being accomplished.

(3 Marks)

(vii) Assuming that it is to be loaded in memory from 0058H, generate the

corresponding addresses for each instruction and write the corresponding

machine code using the operation code table below. (17 Marks)

Mnemonic Opcode

LDS 6C

LDA 00

LDX 04

ADD 18

TIXR B8

STA 0C

JLT 38

S 4

12. (a) Two passes are usually needed to translate an Assembler language program to

machine code; explain briefly what is accomplished in each pass and state the

tables and the variables used in each pass. (5 marks)

(d) Consider the code given below.

000 EXAM START 0

000 FIRST STL RETA 17202000 LDB #LENG 69202 BASE LENGT 000 CLOOP +JSUB RDRE 4B101000 LDA LENGT 03202000 COMP #0 29000001 JEQ ENDFI 33200001 +JSUB WRRE 4B101001 J CLOOP 3F2FE001 ENDFI LDA EOF 03201001 STA BUFFE 0F201002 LDA #3 01000002 STA LENGT 0F200002 +JSUB WRRE 4B101002 J @RET 3E200002 EOF BYTE C’EOF’ 454F4003 RETA RESW 1 003 LENGT RESW 1 003 BUFFE RESB 4096

(j) Some instructions in the code above will change when the program is

loaded in memory from a different memory location other than 0000.

Give the addresses of the instructions which will change and explain

why the others will not change. (3 marks)

(i) Generate the object code for the program above. (12 marks)

13. Consider the program code below written in the simple SIC format.

(a) Modify the program so that it can be assembled in one pass. (10 marks)

1000 EXAM START 1000

1000 LDA ZERO 00104B


1006 LOOP LDX INDEX 041021

1009 LDA ALPHA,X 009024

100C ADD BETA,X 189030

100F STA GAMMA,X 0C903C

1012 LDA INDEX 001021

1015 ADD THREE 181048


101B COMP K300 28104E

101E JLT LOOP 381006

1021 INDEX RESW 1

1024 ALPHA RESW 4

1030 BETA RESW 4

103C GAMMA RESW 4

1048 THREE WORD 3 000003

104B ZERO WORD 0 000000

104E K300 WORD 12 00000C

(b) Use the Load and Go method to show how the above program will appear in

memory. (6 marks)

(c) Explain the following

(i) Programs with Overlay Structures. (2 marks)

(ii) The meaning of the expression: MYNUM EQU * (2 marks)

14. (a) (i) Briefly explain the difference between Program Blocks and Control

Sections. (2 marks)

(ii) What are the directives used to indicate a new Program Block and a

new Control section respectively in a program (2 marks)

(b) Assume that the following piece of code is in a program block called PROGA.

001C LDT NUM 002C NUMBER: RESB 1

(iii) If the beginning address of PROGA is 000A and the code at address

002C is the last directive in PROG A, what is the length of Program

Block PROGA? (2 marks)

(iv) Assume that NUM is defined at address 000E in the next Program

Block PROGB which begins at address 0000H. If the Operation Code

for the instruction LDT is 74, derive the machine instruction for the

Assembler instruction at address 001C in PROGA. (4 marks)

(ci) Explain any 5 Loader Options that can be used to modify the standard

processing of a program. (10 marks)

15. Given below are two control sections PROG1 and PROG2 both loaded in memory.

0000

PROG1 START 0

EXTDEF

EXTREF

0020

LDT NUM1

0027

+STT STORE2

0040

NUM1 RESW 1

0060

RESULT

WORD STORE2 –NUM1

END

0000 PROG2 START 0

EXTDEF

EXTREF

.

0036 +LDT NUM1

003D STT STORE2

0060 STORE2 RESW 1

007C RESULT WORD STORE2 - NUM1

END (a) Give the operands for the Directives EXTREF and EXTDEF in each control

section. (2 marks)

(b) Derive machine code for the instructions at addresses 0020, 0027, 0060 in

PROG1 and 0036, 003D and 007C in PROG2 assuming that the operation

codes for the instructions LDT and STT are 74 and 84 respectively. (6 marks)

(e) Write the object code for the two Control Sections. (8 marks)

(f) Assume that the first control section PROG1 is loaded in memory at address

1200 and that PROG2 is loaded immediately after PROG1, calculate the value

of RESULT (4 marks)



TEST 2

1. Explain the following terms used in Systems Programs:

(i) Assembling

To produce machine language instructions from assembler language along

with information for the loader.

(ii) Loading

Bring the object program into memory for execution.

(iii) Program Relocation

Modifies the object program so that it can be loaded at an address different

from the location originally specified

(iv) Linking

Combines two or more separate object programs and supply the information

needed to allow references between them

(v) Literal Table

A table containing information about the literals. It contains the literal name,

the operand value and the address assigned to the operand when it is placed

in the literal pool

(vi) Literal Origin

It is a directive used to create a literal pool that contains all the literal

operands used since the previous LTORG or the beginning of the program.

2. Briefly explain one of the methods that may be used to assemble a program in

one pass.

One type produces object code directly in memory for immediate execution. No object

code is written out and no loader is needed (load and go). In both cases data item

definitions are put at the top of the program.

The other method produces the usual kind of object program for later execution.

Forward references are entered into lists as before but when the definition of the

symbol is encountered another text record with the correct operand address is

generated. When the program is loaded, this address will be inserted into the

instruction by the action of the loader.

6. Use examples where possible to help you briefly explain the difference between

the following:

�� Program Blocks and Control Sections

A control Section is part of the program that maintains its identity after

assembly; each section can be loaded and relocated independently of each

other. Different control sections are most often used for subroutines or other

logical subdivisions of a program.

Control sections use External References to connect each other.

Program Blocks are segments of code that are re-arranged within a single

object unit.

The assembler rearranges these segments to gather together the pieces of each

block. These blocks are then assigned addresses in the object program with

the blocks appearing in the same order in which they were first began in the

source program.

During pass 1 a separate location counter is maintained for each block. It is

initialized to 0 when the block is first began. At the end of pass 1 the latest

value of the location counter for each block indicates the length of that block.

Control sections differ from program blocks in that they are handled

separately by the assembler

(b) EXTDEFS and EXTREFS

The EXTDEF statement names external that are defined in the current control

section but may also be used by other sections. The EXTREF statement names

symbols used in the current control section but are defined elsewhere.

(c) Absolute expressions and Relative expressions.

An expression that contains only absolute terms (independent of the program

location like constants) is an absolute expression. It may also contain relative

terms so long as the relative terms occur in pairs and the terms in each pair

have opposite signs.

A relative expression is one in which all the relative terms except one can be

paired. The remaining unpaired relative term must have a positive sign.

(d) Literals and immediate operands

A literal is a constant operand whose value is written as part of the instruction

that uses it. The assembler generates the specified value as a constant at some

other memory location. The address of this generated constant is used as the

target address for the machine instruction. An immediate operand is

assembled as part of the machine instruction.

5. Consider the code below

� � � � � � � � �

� � � � � � � � � � � � � �

� � � � � � ! " � #

� � � � � � ! " � #

� � � � � � $ % � & � � � � � �

� � � � ! & '

� � � ' �

� % � ( � ! � � � �

� � $ % � & � ) � � � �

� * % � � � � �

� � � ! � � � � � � � + � , � � � ,

� � � � � � & � � � �

� � � � �

� � � � � � � ! " � #

� � $ % � & � ) � � � �

� � � � � � "

� � - + � , � � � ,

� � � � � � � � � � ) �

� � � � ! " � # � � � ) �

� � � & � � � � � � � � � . �

� � � � & � � ! � � ( & -

� ' � / � � ! � ( & � & � � ! � 0 � & � � � �

a. (i) The Operations codes for the instructions STL, LDB, and COMP

are 14, 68, and 28 respectively, write machine code for the

instructions at addresses 0000, 0003, and 000D.

STL RETADR

Displacement = 0030 – 0003 = 02D

Format = opcode n i x b p e displacement

= 000101 1 1 0 0 1 0 02D = 17202D

LDB #LENGTH

Displacement = 0033 – 0006 = 02D


= 011010 1 0 1 0 1 0 02D = 69202D

COMP #0

Displacement = 000


= 001010 0 1 0 0 0 0 000 = 290000

(ii) What does the Directive at address 1036 mean?

It gives BUFEND the value of the address of the next byte after the

buffer area = 1036

b. Write all the Modification Records that may be needed in this piece of

code. Why don’t the other remaining instructions need Modification

Records?

M^000007^05

M^000014^05

M^000027^05

The other instructions in the program do not need to be modified because in

some cases the operand is not memory or the operand is specified using

program counter relative addressing or base relative addressing.

c. Assume that RDREC and WRREC are in different control sections and

that the variables LENGTH, BUFFER and BUFEND are referenced by

these other control sections:

(i) Insert some extra lines of code in the above program to reflect this

assumption.

These lines are inserted in the program after the first line.

EXTDEF BUFFER, BUFEND, LENGTH

EXTREF RDREC, WRREC

(ii) Write the machine code for the instructions at addresses 0006,

0013 and 0026. The Operation code for the instruction JSUB is 48.

Address = 00000

Format = opcode n i x b p e address

= 010010 1 1 0 0 0 1 00000 = 4B100000

d. Assume that the instruction at address 000A is in a block whose length is

003F and that the operand NUM is at address 0006 of the next block.

Derive the machine code for this instruction.

Effective address = 003F + 0006 = 0045

Displacement = 0045 – 000D = 038


= 0000 00 1 1 0 0 1 0 038 = 0312038

Examinable Question and answer system programming

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