Transcript View File - University of Engineering and Technology, Taxila

Computer Architecture &
Organization
Instructions: PCSpim Tutorial
Engr. Umbreen Sabir
Computer Engineering Department,
University of Engg. & Technology Taxila.
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19/02/2009
CA&O Lecture 05 by Engr. Umbreen Sabir
Adventures in Assembly Land
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What is an Assembler
ASM Directives
ASM Syntax
Intro to PCSPIM
Simple examples
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CA&O Lecture 05 by Engr. Umbreen Sabir
A Simple Programming Task
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Add the numbers 0 to 4 …
10 = 0 + 1 + 2 + 3 + 4
In “C”: int i, sum;
main() {
sum = 0;
for (i=0; i<5; i++)
sum = sum + I; }
Now let’s code it in ASSEMBLY
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CA&O Lecture 05 by Engr. Umbreen Sabir
What IS an Assembler?
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A program for writing programs
 Assembler:
1. A Symbolic LANGUAGE for representing strings of bits.
2. A PROGRAM for translating Assembly Source to binary.
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CA&O Lecture 05 by Engr. Umbreen Sabir
Assembly Source Language
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An Assembly SOURCE FILE contains, in symbolic
text, values of successive bytes to be loaded into
memory... e.g.
.data 0x10000000 Specifies address where following
values are loaded
.byte 1, 2, 3, 4
#First four byte values
.byte 5, 6, 7, 8
# Second four byte values
.word 1, 2, 3, 4
# 4 WORD values (each is 4 bytes)
.asciiz "Comp 411“ # A string of 9 ASCII bytes
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CA&O Lecture 05 by Engr. Umbreen Sabir
Assembly Source Language cont.
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.align 2
# Align to next multiple of 4 (22)
.word 0xfeedbeef # Hex encoded WORD Value
Resulting memory dump:
[0x10000000] 0x04030201 0x08070605 0x00000001 0x00000002
[0x10000010] 0x00000003 0x00000004 0x706d6f43 0x31313420
[0x10000020] 0x00000000 0xfeedbeef 0x00000000 0x00000000
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Notice the byte ordering. This MIPS is “little-endian” (The
least significant byte of a word or half-word has the
lowest address)
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CA&O Lecture 05 by Engr. Umbreen Sabir
Assembler Syntax
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Assembler DIRECTIVES (Keywords prefixed with a
‘.’)
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Control the placement and interpretation of bytes in
memory
.data <addr> Subsequent items are considered data
.text <addr> Subsequent items are considered
instructions
.align N
Skip to next address multiple of 2N
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CA&O Lecture 05 by Engr. Umbreen Sabir
Assembler Syntax cont.
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Allocate Storage
.byte b1, b2, …, bn Store a sequence of bytes (8-bits)
.half h1, h2, …, hn Store a sequence of half-words
(16-bits)
.word w1, w2, …, wn Store a sequence of words (32bits)
.ascii “string”
Stores a sequence of ASCII encoded
bytes
.asciiz “string” Stores a zero-terminated string
.space n
Allocates n successive bytes
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CA&O Lecture 05 by Engr. Umbreen Sabir
Assembler Syntax cont.
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Define scope
.globl sym
Declares symbol to be
visible to other files
.extern sym size Sets size of symbol
defined in another
file (Also makes it
DIRECTLY addressable)
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CA&O Lecture 05 by Engr. Umbreen Sabir
Our Example: Variable Allocation
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Two integer variables (by default 32 bits in MIPS)
.data
.globl sum
.globl i
sum:
.space 4
i:
.space 4
“.data” assembler directive places the following words into the
data segment
“.globl” directives make the “sum” and “i” variables visible to all
other assembly modules.
“.space” directives allocate 4 bytes for each variable
19/02/2009
CA&O Lecture 05 by Engr. Umbreen Sabir
Actual “Code”
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Next we write ASSEMBLY code using the instruction
Mnemonics.
.text 0x80000080
.globl main
Main: add $8,$0,$0
add $9,$0,$0
Loop: addu $8,$8,$9
addi $9,$9,1
slti $10,$9,5
bne $10,$0,loop
end: beq $0,$0,end
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# sum = 0
# for (i = 0; ...
# sum = sum + i;
# for (...; ...; i++
# for (...; i<5;
CA&O Lecture 05 by Engr. Umbreen Sabir
SPIM Simulation
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Let’s tryout our Example
We’ll use the PCSPIM (MIPS backwards)
integrated ASSEMBLER, SIMULATOR, and
DEBUGGER.
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CA&O Lecture 05 by Engr. Umbreen Sabir
Getting Started SPIMing
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The following hints will get you started with SPIM
– By default a small fragment of code is loaded
called the “kernel”. We will discuss the role of this
code in detail, later in the course. Basically it’s
primary job is to branch to the “main” label of your
code. It does so in approximately 4 instructions.
– We will use a raw version of the machine, w/o a
kernel.
– You can “single-step” the machine by pressing
[F10].
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CA&O Lecture 05 by Engr. Umbreen Sabir
Getting Started SPIMing cont.
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The results of the execution of each instruction
are reflected in the register and memory location
values.
Illegal operations result in an “exception” which
causes your code to automatically jump back the
kernel.
For our purposes now, this will be due to a bug in
your program.
Refer to the manual for more fancy usage, such
as setting and executing to breakpoints
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CA&O Lecture 05 by Engr. Umbreen Sabir
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CA&O Lecture 05 by Engr. Umbreen Sabir
“RAW” SPIM
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You’ll need
to change
the default
settings as
follows:
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CA&O Lecture 05 by Engr. Umbreen Sabir
A Slightly More Challenging
Program
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int i, sum;
int a[5] = {7,8,9,10,8};
main() {
sum = 0;
for (i=0; i<5; i++)
sum = sum + a[i]; }
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Once more…let’s encode it in assembly
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Add 5 numbers from a list …
sum = n0 + n1 + n2 + n3 + n4
In “C”:
19/02/2009
CA&O Lecture 05 by Engr. Umbreen Sabir
Variable Allocation
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We cheated in our last example. Generally, variables
will be allocated to memory locations, rather than
registers.
This time we add the contents of an array
.data
.extern sum 4
.extern i 4
.extern a 20
sum: .space 4
i:
.space 4
a: .word 7,8,9,10,8
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CA&O Lecture 05 by Engr. Umbreen Sabir
Variable Allocation cont.
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“.word” allows us to initialize a list of
sequential words in memory. The label
represents the address of the first word in the
list, or the name of the array
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CA&O Lecture 05 by Engr. Umbreen Sabir
The New Code
.text 0x80000080
.globl main
main:
sw $0,sum
sw $0,i
lw $9,i
lw $8,sum
loop:
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# sum = 0;
# for (i = 0;
# allocate register for i
# allocate register for sum
19/02/2009
CA&O Lecture 05 by Engr. Umbreen Sabir
The New Code
sll $10,$9,2
lw $10,a($10)
addu $8,$8,$10
sw $8,sum
addi $9,$9,1
sw $9,i
slti $10,$9,5
bne $10,$0,loop
end: beq $0,$0,end
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# covert "i" to word offset
# load a[i]
# sum = sum + a[i];
# update variable in memory
# for (...; ...; i++
# update memory
# for (...; i<5;
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CA&O Lecture 05 by Engr. Umbreen Sabir
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CA&O Lecture 05 by Engr. Umbreen Sabir
A Coding Challenge
What is the largest Fibonacci number less
than 100?
 Fibonacci numbers:
Fi+1 = Fi + Fi-1
F0 = 0
F1 = 1
 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …
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CA&O Lecture 05 by Engr. Umbreen Sabir
A Coding Challenge
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In “C”
int x, y;
main() {
x = 0; y = 1;
while (y < 100) {
int t = x;
x = y;
y = t + y;}
}
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CA&O Lecture 05 by Engr. Umbreen Sabir
Next Lecture and Reminders
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Next lecture
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MIPS ISA
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Assignment Due – 20 Feb 2009
19/02/2009
CA&O Lecture 05 by Engr. Umbreen Sabir
END OF LECTURE 5
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CA&O Lecture 05 by Engr. Umbreen Sabir