Post on 12-Dec-2020
Ibrahim Hazmi | ECE 255 LAB
Lab1: Using ECLIPS Environment
ECE 255 - iHaz - 2019
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ECE 255 LAB Lab Introduction to Computer Architecture
Ibrahim Hazmi | ECE 255 LAB
IBRAHIMHAZMI
Contact me at:Email: ihaz@uvic.ca
My UVic Blog:https://onlineacademiccommunity.uvic.ca/ihaz/
CENG 255 B01 TA
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Ibrahim Hazmi | ECE 255 LAB
1 What is a cross-assembler? (0.5)An assembler that converts mnemonics to binary opcodes, other than the host machine’s2 What is an exception? (0.5)
A change of control flow from the current program execution due to errors, interrupts, … etc.4 Comment on the program in 1.4.2 (1) (What it does) (0.5)5 Comment on the program in 1.4.3 (1) (What it does) (0.5)
Pre-Lab Answers
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Ibrahim Hazmi | ECE 255 LAB
OpCode!
Add/Sub0 0 0 1 1 0 I Rn Rs Rd
0 0 0 1 1 0 0R5 R4 R0
1 0 1 1 0 0 0 0 01 9 6 0
0 0 0 1 1 0 1R5 R4 R0
1 0 1 1 0 0 0 0 01 B 6 0
Add Rd, Rs, Rn (I=0) Sub Rd, Rs, Rn (I=1)
Move/Cmp/Add/Sub0 0 1 Op Rd Offset8/Immediate
0 0 1Mov R0 50 0 0 0 0 0 0 0 0 0 1 0 1
2 0 0 5
0 0 1Cmp R0 50 1 0 0 0 0 0 0 0 0 1 0 1
2 8 0 5
Mov Rd, offset (Op=00) Cmp Rd, Rn/Offset (Op=01)Add Rd, Rs, Rn (Op=10) Sub Rd, Rs, Rn (Op=11)
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Hex value for ‘sub r0, r4, r5’ ? (Op = 1)
Ibrahim Hazmi | ECE 255 LAB 5
https://www.ece.uvic.ca/~ece255/lab/
LAB WEB-PAGE
Ibrahim Hazmi | CENG 255 LAB Ibrahim Hazmi | ECE 255 LAB
Part1
Import an existing project to Eclipse View and Read Memory after downloading the executable binary file to the ARM board. Change the content of the Memory with address “0x2000146” to a “SUBTRACT”
Part2Describe the function of the .asm Program Build and Debug the project, then Examine the contents of the Memory and Registers
Part3
Program function and listingMemory & Registers before & after executionChange the program for 32 elements, and Check for limitation
The Lab Task
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Ibrahim Hazmi | ECE 255 LAB
Lab Hardware Settings!
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Before starting our lab, let’s explore Assembly
Ibrahim Hazmi | CENG 255 LAB Ibrahim Hazmi | ECE 255 LAB
How the Assembly code looks
}
{___
{
__ __ __
__ __ __ __
Comments
Directives (Assembler)
Instructions
Label
} Operands________
____
Now let’s get back to instruction types
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Ibrahim Hazmi | CENG 255 LAB Ibrahim Hazmi | ECE 255 LAB
How Debug mode looks ?
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Ibrahim Hazmi | ECE 255 LAB
Instruction Types
ALU & REG
ADD/SUB
CMP
AND/OR
MEM & REG
MOV
STORE (ST)
LOAD (LD)
PROG. FLPW
BRANCH
BL
BE
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Ibrahim Hazmi | ECE 255 LAB
How instructions are executed in ARM
Reg Content
r0r1r2r3r4r5r6r7
SPLRPC
Address Data
RAM
CPSRN Z C V
Regiters
ALUN
ZC
OOp
n1 n2
Result
data
instructions
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Ibrahim Hazmi | ECE 255 LAB
How instructions are executed in ARM
Reg Content
r0r1 00001000
r2r3r4r5r6r7
SPLRPC 2000AA02
Address Data
00001000 1500001004 10
2000AA00 ldr r1, =l1
2000AA02 ldr r2, =l2
2000AA04 ldr r3, [r1]
2000AA06 ldr r4, [r2]
2000AA08 add r0, r4, r3
2000AA0A sub r0, r4, r3
2000AA0C str r0, [r1]
2000AA0E
2000XXXX xxxxxx
2000XXXX xxxxxx
RAMCPSRN Z C V
Regiters
L1L2
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Assembly Codeldr ldr ldr ldr
add sub str
r1, =l1 r2, =l2 r3, [r1] r4, [r2]
r0, r4, r3 r0, r4, r3 r0, [r1]
Ibrahim Hazmi | ECE 255 LAB
How instructions are executed in ARM
Reg Content
r0r1 00001000
r2 00001004
r3r4r5r6r7
SPLRPC 2000AA04
Address Data
00001000 1500001004 10
2000AA00 ldr r1, =l1
2000AA02 ldr r2, =l2
2000AA04 ldr r3, [r1]
2000AA06 ldr r4, [r2]
2000AA08 add r0, r4, r3
2000AA0A sub r0, r4, r3
2000AA0C str r0, [r1]
2000AA0E
2000XXXX xxxxxx
2000XXXX xxxxxx
RAM
Assembly Codeldr ldr ldr ldr
add sub str
r1, =l1 r2, =l2 r3, [r1] r4, [r2]
r0, r4, r3 r0, r4, r3 r0, [r1]
CPSRN Z C V
Regiters
L1L2
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Ibrahim Hazmi | ECE 255 LAB
How instructions are executed in ARM
Reg Content
r0r1 00001000
r2 00001004
r3 15r4r5r6r7
SPLRPC 2000AA06
Address Data
00001000 1500001004 10
2000AA00 ldr r1, =l1
2000AA02 ldr r2, =l2
2000AA04 ldr r3, [r1]
2000AA06 ldr r4, [r2]
2000AA08 add r0, r4, r3
2000AA0A sub r0, r4, r3
2000AA0C str r0, [r1]
2000AA0E
2000XXXX xxxxxx
2000XXXX xxxxxx
RAM
Assembly Codeldr ldr ldr ldr
add sub str
r1, =l1 r2, =l2 r3, [r1] r4, [r2]
r0, r4, r3 r0, r4, r3 r0, [r1]
CPSRN Z C V
Regiters
L1L2
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Ibrahim Hazmi | ECE 255 LAB
How instructions are executed in ARM
Reg Content
r0r1 00001000
r2 00001004
r3 15r4 10r5r6r7
SPLRPC 2000AA08
Address Data
00001000 1500001004 10
2000AA00 ldr r1, =l1
2000AA02 ldr r2, =l2
2000AA04 ldr r3, [r1]
2000AA06 ldr r4, [r2]
2000AA08 add r0, r4, r3
2000AA0A sub r0, r4, r3
2000AA0C str r0, [r1]
2000AA0E
2000XXXX xxxxxx
2000XXXX xxxxxx
RAM
Assembly Codeldr ldr ldr ldr
add sub str
r1, =l1 r2, =l2 r3, [r1] r4, [r2]
r0, r4, r3 r0, r4, r3 r0, [r1]
CPSRN Z C V
Regiters
L1L2
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Ibrahim Hazmi | ECE 255 LAB
How instructions are executed in ARM
Reg Content
r0 25r1 00001000
r2 00001004
r3 15r4 10r5r6r7
SPLRPC 2000AA0A
Address Data
00001000 1500001004 10
2000AA00 ldr r1, =l1
2000AA02 ldr r2, =l2
2000AA04 ldr r3, [r1]
2000AA06 ldr r4, [r2]
2000AA08 add r0, r4, r3
2000AA0A sub r0, r4, r3
2000AA0C str r0, [r1]
2000AA0E
2000XXXX xxxxxx
2000XXXX xxxxxx
RAM
Assembly Codeldr ldr ldr ldr
add sub str
r1, =l1 r2, =l2 r3, [r1] r4, [r2]
r0, r4, r3 r0, r4, r3 r0, [r1]
CPSRN Z C V0 0 0 0
Regiters
L1L2{
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ALUN
ZC
O Sub
n1 n2
Result
Ibrahim Hazmi | ECE 255 LAB
How instructions are executed in ARM
Reg Content
r0 -5r1 00001000
r2 00001004
r3 15r4 10r5r6r7
SPLRPC 2000AA0C
Address Data
00001000 1500001004 10
2000AA00 ldr r1, =l1
2000AA02 ldr r2, =l2
2000AA04 ldr r3, [r1]
2000AA06 ldr r4, [r2]
2000AA08 add r0, r4, r3
2000AA0A sub r0, r4, r3
2000AA0C str r0, [r1]
2000AA0E
2000XXXX xxxxxx
2000XXXX xxxxxx
RAM
Assembly Codeldr ldr ldr ldr
add sub str
r1, =l1 r2, =l2 r3, [r1] r4, [r2]
r0, r4, r3 r0, r4, r3 r0, [r1]
CPSRN Z C V1 0 0 0
Regiters
L1L2
#517
ALUN
ZC
O Sub
n1 n2
Result
Ibrahim Hazmi | ECE 255 LAB
How instructions are executed in ARMAddress Data
00001000 -5 (FFFFFFFB)00001004 10 (0000000A)
2000AA00 ldr r1, =l1
2000AA02 ldr r2, =l2
2000AA04 ldr r3, [r1]
2000AA06 ldr r4, [r2]
2000AA08 add r0, r4, r3
2000AA0A sub r0, r4, r3
2000AA0C str r0, [r1]
2000AA0E
2000XXXX xxxxxx
2000XXXX xxxxxx
RAM
Assembly Codeldr ldr ldr ldr
add sub str
r1, =l1 r2, =l2 r3, [r1] r4, [r2]
r0, r4, r3 r0, r4, r3 r0, [r1]
L1L2
}
Reg Content
r0 -5r1 00001000
r2 00001004
r3 15r4 10r5r6r7
SPLRPC 2000AA0E
CPSRN Z C V1
Regiters
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Ibrahim Hazmi | CENG 255 LAB Ibrahim Hazmi | ECE 255 LAB
Part1
Import an existing project to Eclipse View and Read Memory after downloading the executable binary file to the ARM board. Change the content of the Memory with address “0x2000146” to a “SUBTRACT”
Part2Describe the function of the .asm Program Build and Debug the project, then Examine the contents of the Memory and Registers
Part3
Program function and listingMemory & Registers before & after executionChange the program for 32 elements, and Check for limitation
The Lab Task
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Ibrahim Hazmi | CENG 255 LAB Ibrahim Hazmi | ECE 255 LAB
Lab 1as
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Ibrahim Hazmi | CENG 255 LAB Ibrahim Hazmi | ECE 255 LAB
Lab 1bs21
Ibrahim Hazmi | CENG 255 LAB Ibrahim Hazmi | ECE 255 LAB
1
Include a well-commented listing of your program. Comments should include register usage (i.e., which variables are kept in which registers), and a description of all symbols
2 Snap shots of the data section before and after execution of the program
3 Answer all questions in the lab manual!
Deliverable
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1 Remember to backup your code always!
2 Your assembly code should start with defining your data and its space!
Advices for the lab work
Ibrahim Hazmi | CENG 255 LAB Ibrahim Hazmi | ECE 255 LAB 23
A Quick Flowchartfor How to work on
Eclipse
Ibrahim Hazmi | ECE 255 LAB
References
https://www.youtube.com/watch?v=cNN_tTXABUA
https://www.youtube.com/watch?v=AVy9LLPTH9Q
http://www.ece.uvic.ca/~kinli/ceng255/ARM7-TDMI-manual-pt3.pdf
https://cs.wellesley.edu/~cs110/lectures/compModel/computer.html#fig:computer-labels
https://niteshprasad.wordpress.com/2014/04/20/lesson-1-how-central-processing-unit-works/
http://www.ece.uvic.ca/~kinli/ceng255/PM0215_STM32F0xxx%20Cortex_M0%20programming%20manual-commented.pdf
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