Computer Science 210 s1c Computer Systems 1 2014 Semester 1 Lecture Notes James Goodman (revised by...

Computer Science 210 s1c

Computer Systems 12014 Semester 1

Lecture Notes

James Goodman (revised by Robert Sheehan)

Credits: “McGraw-Hill” slides prepared by Gregory T. Byrd, North Carolina State University

Machine language programs

Lecture 10

1

2

Using Branch Instructions

Compute sum of 12 integers.Numbers start at location x3100. Program starts at location x3000.

R1 x3100R3 0

R2 12

R2=0?

R4 M[R1]R3 R3+R4R1 R1+1R2 R2-1

NO

YES

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Sample Program

3

Address Instruction Comments

x3000 1 1 1 0 0 0 1 0 1 1 1 1 1 1 1 1 R1 x3100 (PC+0x0FF)

x3001 0 1 0 1 0 1 1 0 1 1 1 0 0 0 0 0 R3 0

x3002 0 1 0 1 0 1 0 0 1 0 1 0 0 0 0 0 R2 0

x3003 0 0 0 1 0 1 0 0 1 0 1 0 1 1 0 0 R2 12

x3004 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 If Z, goto x300A (PC+5)

x3005 0 1 1 0 1 0 0 0 0 1 0 0 0 0 0 0 Load next value to R4

x3006 0 0 0 1 0 1 1 0 1 1 0 0 0 1 0 0 Add to R3

x3007 0 0 0 1 0 0 1 0 0 1 1 0 0 0 0 1 Increment R1 (pointer)

x3008 0 0 0 1 0 1 0 0 1 0 1 1 1 1 1 1 Decrement R2 (counter)

x3009 0 0 0 0 1 1 1 1 1 1 1 1 1 0 1 0 Goto x3004 (PC-6)

4

JMP (Register)

Jump is an unconditional branch -- always taken. Target address is the contents of a register. Allows any target address.


We could have used this as the last instruction in the previous sample program, but we would have had to put the destination address into a register first.

5

TRAP

Calls a service routine, identified by 8-bit “trap vector.”

When routine is done, PC is set to the instruction following TRAP.(We’ll talk about how this works later.)

vector routine

x23 input a character from the keyboard

x21 output a character to the monitor

x25 halt the program


6

Another Example

Count the occurrences of a character in a file Program begins at location x3000 Read character from keyboard Load each character from a “file”

• File is a sequence of memory locations• Starting address of file is stored in the memory location

immediately after the program If file character equals input character, increment counter End of file is indicated by a special ASCII value: EOT (x04) At the end, print the number of characters and halt

(assume there will be less than 10 occurrences of the character)

A special character used to indicate the end of a sequenceis often called a sentinel.

Useful when you don’t know ahead of time how many timesto execute a loop.


5-8

Program (1 of 2)


x3000 0 1 0 1 0 1 0 0 1 0 1 0 0 0 0 0 R2 0 (counter)

x3001 0 0 1 0 0 1 1 0 0 0 0 1 0 0 0 0 R3 M[x3102] (ptr)

x3002 1 1 1 1 0 0 0 0 0 0 1 0 0 0 1 1 Input to R0 (TRAP x23)

x3003 0 1 1 0 0 0 1 0 1 1 0 0 0 0 0 0 R1 M[R3]

x3004 0 0 0 1 1 0 0 0 0 1 1 1 1 1 0 0 R4 R1 – 4 (EOT)

x3005 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 If Z, goto x300E

x3006 1 0 0 1 0 0 1 0 0 1 1 1 1 1 1 1 R1 NOT R1

x3007 0 0 0 1 0 0 1 0 0 1 1 0 0 0 0 1 R1 R1 + 1

X3008 0 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 R1 R1 + R0

x3009 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 1 If N or P, goto x300B

5-9

Program (2 of 2)


x300A 0 0 0 1 0 1 0 0 1 0 1 0 0 0 0 1 R2 R2 + 1

x300B 0 0 0 1 0 1 1 0 1 1 1 0 0 0 0 1 R3 R3 + 1

x300C 0 1 1 0 0 0 1 0 1 1 0 0 0 0 0 0 R1 M[R3]

x300D 0 0 0 0 1 1 1 1 1 1 1 1 0 1 1 0 Goto x3004

x300E 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 R0 M[x3013]

x300F 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 R0 R0 + R2

x3010 1 1 1 1 0 0 0 0 0 0 1 0 0 0 0 1 Print R0 (TRAP x21)

x3011 1 1 1 1 0 0 0 0 0 0 1 0 0 1 0 1 HALT (TRAP x25)

X3012 Starting Address of File

x3013 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 ASCII x30 (‘0’)

10

Filled arrow = info to be processed.

Unfilled arrow= control signal.

LC-3 Data PathRevisited

11

Data Path Components

Global bus special set of wires that carry

a 16-bit signal to many components

inputs to the bus are “tri-state devices,” that only place a signal on the bus when they are enabled

only one (16-bit) signal should be enabled at any time

• control unit decides which signal “drives” the bus

any number of components can read the bus

• register only captures bus data if it is write-enabled by the control unit

Memory Control and data registers

for memory and I/O devices memory: MAR, MDR (also

control signal for read/write)


5-12


ALU Accepts inputs from register file

and from sign-extended bits from IR (immediate field). Output goes to bus.

• used by condition code logic, register file, memory

Register File Two read addresses (SR1, SR2), one write address (DR) Input from bus

• result of ALU operation or memory read Two 16-bit outputs

• used by ALU, PC, memory address• data for store instructions passes through ALU

5-13


PC and PCMUX Three inputs to PC, controlled by PCMUX

1. PC+1 – FETCH stage2. Address adder – BR, JMP3. bus – TRAP (discussed later)

MAR and MARMUX Two inputs to MAR, controlled by MARMUX

1. Address adder – LD/ST, LDR/STR2. Zero-extended IR[7:0] -- TRAP (discussed later)

5-14


Condition Code Logic Looks at value on bus and generates N, Z, P signals Registers set only when control unit enables them (LD.CC)

• only certain instructions set the codes(ADD, AND, NOT, LD, LDI, LDR, LEA)

Control Unit – Finite State Machine On each machine cycle, changes control signals for next

phaseof instruction processing• who drives the bus? (GatePC, GateALU, …)• which registers are write enabled? (LD.IR, LD.REG, …)• which operation should ALU perform? (ALUK)• …

Logic includes decoder for opcode, etc.



Lecture Notes


Credits: Slides prepared by Gregory T. Byrd, North Carolina State University

Chapter 7: Assembly Language

15

Lecture 11

16

Human-Readable Machine Language

Computers like ones and zeros…

Humans like symbols…

Assembler is a program that turns symbols intomachine instructions.

ISA-specific:close correspondence between symbols and instruction set• mnemonics for opcodes• labels for memory locations

additional operations for allocating storage and initializing data

ADD R6,R2,R6 ; increment index reg.

0001110010000110


17

LC-3 Assembly Language Syntax

Each line of a program is one of the following: an instruction an assembler directive (or pseudo-op) a comment

Whitespace (between symbols) and case are ignored.Comments (beginning with “;”) are also ignored.

An instruction has the following format:

LABEL OPCODE OPERANDS ; COMMENTS

optional mandatory


18

Opcodes and Operands

Opcodes reserved symbols that correspond to LC-3 instructions listed in Appendix A

• ex: ADD, AND, LD, LDR, …

Operands registers -- specified by Rn, where n is the register number numbers -- indicated by # (decimal) or x (hex) or b (binary) label -- symbolic name of memory location separated by comma number, order, and type correspond to instruction format

• ex:ADD R1,R1,R3ADD R1,R1,#3LD R6,NUMBERBRz LOOP


19

Labels and Comments

Label placed at the beginning of the line assigns a symbolic name to the address corresponding to

line• ex:

LOOP ADD R1,R1,#-1BRp LOOP

Comment anything after a semicolon is a comment ignored by assembler used by humans to document/understand programs tips for useful comments:

• avoid restating the obvious, as “decrement R1”• provide additional insight, as in “accumulate product in

R6”• use comments to separate pieces of program


20

Assembler Directives

Pseudo-operations do not refer to operations executed by program used by assembler look like instruction, but “opcode” starts with a full stop

Opcode Operand Meaning

.ORIG address starting address of program

.END end of program

.BLKW n allocate n words of storage

.FILL n allocate one word, initialize with value n

.STRINGZ n-character

string

allocate n+1 locations, initialize w/characters and null terminator


Hello World in LC-3 Assembler

; Hello world in LC-3 assembler; I could have just used "PUTS" but that isn't fun.

.ORIG x3000LEA R1, hello ; R1 points to next character

loop LDR R0, R1, #0 ; R0 holds next characterBRz finishTRAP x21 ; or just OUT prints R0[7:0]ADD R1, R1, #1BRnzp loop

finish TRAP x25 ; or HALT

hello .STRINGZ "Hello world".END

21

22

Trap Codes

LC-3 assembler provides “pseudo-instructions” foreach trap code, so you don’t have to remember them.

Code Equivalent Description

HALT TRAP x25 Halt execution and print message to console.

IN TRAP x23 Print prompt on console,read (and echo) one character from keybd.Character stored in R0[7:0].

OUT TRAP x21 Write one character (in R0[7:0]) to console.

GETC TRAP x20 Read one character from keyboard.Character stored in R0[7:0].

PUTS TRAP x22 Write null-terminated string to console.Address of string is in R0.


23

Style Guidelines

Use the following style guidelines to improvethe readability and understandability of your programs:1. Provide a program header, with author’s name, date, etc.,

and purpose of program. 2. Start labels, opcode, operands, and comments in same

columnfor each line. (Unless entire line is a comment.)

3. Use comments to explain what each register does.4. Give explanatory comment for most instructions.5. Use meaningful symbolic names.

• Mixed upper and lower case for readability.• ASCIItoBinary, InputRoutine, SaveR1

6. Provide comments between program sections.7. Each line must fit on the page -- no wraparound or

truncations.• Long statements split in aesthetically pleasing manner.


24

Sample Program

Count the occurrences of a character in a file.


TEST:

BRz OUTPUT

OUTPUT:

25

; Program to count occurrences of a character in a file.; Character to be input from the keyboard.; Result to be displayed on the monitor.; Program only works if no more than 9 occurrences are found.; ;; Initialization;

.ORIG x3000AND R2, R2, #0 ; R2 is counter, initially 0LD R3, PTR ; R3 is pointer to charactersGETC ; R0 gets character inputLDR R1, R3, #0 ; R1 gets first character

;; Test character for end of file;TEST ADD R4, R1, #-4 ; Test for EOT (ASCII x04)

BRz OUTPUT ; If done, prepare the output;; Test character for match. If a match, increment count.;

NOT R1, R1ADD R1, R1, R0 ; If match, R1 = xFFFFNOT R1, R1 ; If match, R1 = x0000BRnp GETCHAR ; If no match, do not incrementADD R2, R2, #1

;; Get next character from file.;GETCHAR ADD R3, R3, #1 ; Point to next character.

LDR R1, R3, #0 ; R1 gets next char to testBRnzp TEST

;; Output the count.;OUTPUT LD R0, ASCII ; Load the ASCII template

ADD R0, R0, R2 ; Convert binary count to ASCIIOUT ; ASCII code in R0 is displayed.HALT ; Halt machine

;; Storage for pointer and ASCII template;ASCII .FILL x0030PTR .FILL x4000

.END

Do this

Download the LC‐3 simulator package from the resources page<http://www.cs.auckland.ac.nz/compsci210s1c/resources/>. For running on Windows, read the document LC3WinGuide.pdf. (You may run the simulator under Linux: read the document LC3_unix.pdf).

Follow the instructions for running a program, creating the files described in the example and execute the program.

Create a source file from the text of program discussed in the lecture (figures 5.16 & 7.2 in the book).

Create a “file” starting in the memory at location x4000.Assemble the programme.Execute the programme, typing different characters and make

sure the programme prints the correct result.What goes wrong if the character you enter occurs more than 10

times in the file?

26



Lecture Notes


Credits: Slides prepared by Gregory T. Byrd, North Carolina State University

27

Lecture 12

The Assembly Process

28

First Pass: Constructing the Symbol Table

1. Find the .ORIG statement,which tells us the address of the first instruction.• Initialize location counter (LC), which keeps track of

thecurrent instruction.

2. For each non-empty line in the program:a) If line contains a label, add label and LC to symbol

table.b) Increment LC.

– NOTE: If statement is .BLKW or .STRINGZ,increment LC by the number of words allocated.

3. Stop when .END statement is reached.

NOTE: A line that contains only a comment is considered an empty line.


29

Symbol Table Construction

Construct the symbol table for the program in Figure 7.1 .

Symbol Address


30











.END

0x30000x3001 PTR

31



Symbol AddressPTR ?


32











.END

0x30000x30010x30020x3003

0x3004TEST

33



Symbol AddressPTR ?


TEST 0x3004

34











.END

0x30000x30010x30020x3003

0x30040x3005 OUTPUT

35



Symbol AddressPTR ?

TEST 0x3004


OUTPUT ?

36











.END

0x30000x30010x30020x3003

0x30040x3005

0x30060x30070x30080x3009 GETCHAR

37



Symbol AddressPTR ?

TEST 0x3004

OUTPUT ?


GETCHAR ?

38











.END

0x30000x30010x30020x3003

0x30040x3005

0x30060x30070x30080x30090x300A

0x300BGETCHAR

39



Symbol AddressPTR ?

TEST 0x3004

OUTPUT ?

GETCHAR ?


0x300B

40











.END

0x30000x30010x30020x3003

0x30040x3005

0x30060x30070x30080x30090x300A

0x300B0x300C0x300D TEST

41



Symbol AddressPTR ?

TEST 0x3004

OUTPUT ?

GETCHAR 0x300B


42



Symbol AddressPTR ?

TEST 0x3004

OUTPUT ?

GETCHAR 0x300B


43











.END

0x30000x30010x30020x3003

0x30040x3005

0x30060x30070x30080x30090x300A

0x300B0x300C0x300D

0x300EOUTPUT

44



Symbol AddressPTR ?

TEST 0x3004

OUTPUT ?

GETCHAR 0x300B


0x300E

45











.END

0x30000x30010x30020x3003

0x30040x3005

0x30060x30070x30080x30090x300A

0x300B0x300C0x300D

0x300E ASCII

46



Symbol AddressPTR ?

TEST 0x3004

OUTPUT 0x300E

GETCHAR 0x300B


ASCII ?

47











.END

0x30000x30010x30020x3003

0x30040x3005

0x30060x30070x30080x30090x300A

0x300B0x300C0x300D

0x300E0x300F0x30100x3011

0x3012ASCII

48



Symbol AddressPTR ?

TEST 0x3004

OUTPUT 0x300E

GETCHAR 0x300B

ASCII ?


0x3012

49











.END

0x30000x30010x30020x3003

0x30040x3005

0x30060x30070x30080x30090x300A

0x300B0x300C0x300D

0x300E0x300F0x30100x3011

0x30120x3013PTR

50



Symbol AddressPTR ?

TEST 0x3004

OUTPUT 0x300E

GETCHAR 0x300B

ASCII 0x3012


0x3013

51

Second Pass: Generating Machine Language

For each executable assembly language statement,generate the corresponding machine language instruction.

If operand is a label,look up the address from the symbol table.

Potential problems: Improper number or type of arguments

• ex: NOT R1,#7ADD R1,R2ADD R3,R3,NUMBER

Immediate argument too large• ex: ADD R1,R2,#1023

Address (associated with label) more than 256 from instruction

• can’t use PC-relative addressing mode


52

Practice

Using the symbol table constructed earlier,translate these statements into LC-3 machine language.


Statement Machine LanguageLD R3,PTR

ADD R4,R1,#-4

LDR R1,R3,#0

BRnp GETCHAR

53

Practice

Using the symbol table constructed earlier,translate these statements into LC-3 machine language.


54

LC-3 Assembler

Using “assemble” (Unix) or LC3Edit (Windows),generates several different output files.

This one getsloaded into thesimulator.


55

Object File Format

LC-3 object file contains Starting address (location where program must be loaded),

followed by… Machine instructions

Example Beginning of “count character” object file looks like this:

0011000000000000010101001010000000100110000100011111000000100011

.

.

.

.ORIG x3000

AND R2, R2, #0

LD R3, PTR

TRAP x23


56

Multiple Object Files

An object file is not necessarily a complete program.

system-provided library routines code blocks written by multiple developers

For LC-3 simulator, can load multiple object files into memory,then start executing at a desired address.

system routines, such as keyboard input, are loaded automatically• loaded into “system memory,” below x3000• user code should be loaded between x3000 and xFDFF

each object file includes a starting address be careful not to load overlapping object files


57

Linking and Loading

Loading is the process of copying an executable imageinto memory.

more sophisticated loaders are able to relocate imagesto fit into available memory

must readjust branch targets, load/store addresses

Linking is the process of resolving symbols betweenindependent object files.

suppose we define a symbol in one module,and want to use it in another

some notation, such as .EXTERNAL, is used to tell assembler that a symbol is defined in another module

linker will search symbol tables of other modules to resolve symbols and complete code generation before loading


58

News from the NYTimes (18 June 1996)

“When a computer runs out of [main memory], modern operating systems automatically use the memory on the hard drive. But today’s hard drives retrieve data at speeds of about 10 milliseconds (millionths of a second). That seems fast until you consider that modern RAM can do this at 60 nanoseconds (billionths of a second), more than 150 times as fast.”

What’s wrong with this statement??

59

“Correction: June 19, 1996, Wednesday The Personal Computers column in Science Times yesterday, about options for increasing computer memory, incorrectly described milliseconds (they are thousandths of a second, not millionths) and therefore misstated the difference in speed between random access memory (RAM) and hard drives. RAM can be more than 100,000 times, not 150 times, as fast.”

60

Time Line

10-10 10-7 10010-9 10-8 10-6 10-5 10-4 10-3 10-2 10-1

Mill

isec

ond

Mic

rose

cond

Nan

osec

ond

Seco

nd

Time (Logarithmic Scale)

1 month1 day1 hour1 minute1 second 1 year

Mic

roCen

tury

Nan

oCen

tury

Cen

tury

Mill

iCen

tury

Scale by 31,557,600

61

Speed Line

Time for light to travel 30 cm

One clock period2 GHz

Total Diskaccess time

Cache miss time(Memory access time)

Cache hit time

Executeone

instruction

(best case)

Time for sound to travel 30 cm

One disk revolution(6-8 ms)

Transfer 1 char at 56K

baud

Read 1 byte from disk

10-10 10-7 10010-9 10-8 10-6 10-5 10-4 10-3 10-2 10-1

Mill

isec

ond

Mic

rose

cond

Nan

osec

ond

Seco

nd

Time (Logarithmic Scale)

Computer Science 210 s1c Computer Systems 1 2014 Semester 1 Lecture Notes James Goodman (revised by...

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