1 Copyright 1998 Morgan Kaufmann Publishers, Inc. All rights reserved. MIPS Assembly language In...

1Copyright 1998 Morgan Kaufmann Publishers, Inc. All rights reserved.

MIPS Assembly language

In computer programs we have :

DataInstructions (Arithmetic & control)

We need a memory & a CPU:


What should the CPU do?

• We need to transfer data from registers to the memory and vice versa

• We need to perform arithmetic operations on the data residing in registers

• We need to have the ability of controlling the flow of the program (Ifs and Jumps)


The machine understands only bits!


Humans prefer a language “closer” to English


Labels will improve the situation


High-level language is even better


The process of Compilation


Preface

We’ll learn:

• How a computer is built

• Some performance analysis

• Advanced subject (Pipeline, caches) that are used in all modern computers

The course book:

Computer Organization & Design The hardware/software interface, David A. Patterson and John L. Hennessey.

Second Edition 1998


Machine language, Assembly and C

swap(int v[], int k){int temp; temp = v[k]; v[k] = v[k+1]; v[k+1] = temp;}

swap: muli $2, $5,4 add $2, $4,$2 lw $15, 0($2) lw $16, 4($2) sw $16, 0($2) sw $15, 4($2) jr $31

00000000101000010000000000011000000000001000111000011000001000011000110001100010000000000000000010001100111100100000000000000100101011001111001000000000000000001010110001100010000000000000010000000011111000000000000000001000

Binary machinelanguageprogram(for MIPS)

C compiler

Assembler

Assemblylanguageprogram(for MIPS)

High-levellanguageprogram(in C)

• The CPU understands machine Language only

• Assembly language is easier to understand:

- Abstraction

- A unique translation

• C language (a high level language):

- The translation is

not unique !

- It depends on the Compiler

and the optimization requested

- It is portable (fits every CPU)

When do we use Assembly?

C is closer to regular English.( The high level language can be adjusted to the desired processing). High productivity: In a single sentence we define many machine operations. And the most important: It does not depend on the CPU !!!

Looks like regular English. Depends on the specific CPU. Every CPU has a different set of Assembly instructions since it has different sets of registers and machine operations.

Nowadays we use Assembly only when we have to:* When processing time is critical and we need “manually” adjustments of the code* When all resources of the CPU (e.g., Overflow, Index registers etc.) are needed, but not supported by the hig level language* When memory is critical an an optimization of its management is required (e.g., in DSPs)


Instruction Set Architecture

• There is similarity between Assembly languages of different CPUs

• We’ll learn the MIPS Assembly language. It was developed in the 80’s and was used in workstation such as Silicon Graphics, NEC, Sony.

• RISC v. CISC

– Reduced Instruction Set Computer - MIPS– 8086 - Complex Instruction Set Computer

Our motto is: “More is less”

By that we mean that a smaller set of simpler instructions results with a better performance. This is so, since the h/w required is simpler, it is therefore faster and takes less Silicon space, which means less expensive.


1st design rule: Simplicity favors Regularity

Arithmetic operations:Arithmetic operations:

• MIPS

addi a,b,100 # a=b+100

add a,b,c # a=b+c• 8086

ADD EAX,B # EAX= EAX+B

We prefer a simple mechanism with minimalistic instructions such as: R3 = R1 op R2 on a CPU which might be easier to program since it has instructions like: R5 = ( R1 op1 R2) op2 (R3 op3 R4 )but is much harder to design and implement


2nd design rule: Smaller is faster

• Arithmetic operations are allowed only on registers• The operands could be registers or a single constant • There are only 32 registers

( spilling is applied when needed) • A register contains a word = 32 bits = 4 bytes• We use conventions

$1,$2 …

$s0,$s1 ... - C variables

$t1,$t2 … - temporary vars.

An example:

f=(g+h)-(k+j) # $s0=f, $s1=g, $s2=h, $s3=k, $s4=j

add $t0,$s1,$s2

add $t1,$s3,$s4

sub $s0, $t0, $t1

The rgisters are denoted $0 - $31 or by names related to their job. There is a convention of their jobs.

The example describes how the sentence: f=(g+h)-(k+j) is translated into Assembly language instructions.The registers are the heart of the CPU. Accessing registers is faster than accessing the memory. We access 3 registers simultaneously: read from two and write to the 3rd.

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Policy of Use Conventions

Name Register number Usage$zero 0 the constant value 0$v0-$v1 2-3 values for results and expression evaluation$a0-$a3 4-7 arguments$t0-$t7 8-15 temporaries$s0-$s7 16-23 saved$t8-$t9 24-25 more temporaries$gp 28 global pointer$sp 29 stack pointer$fp 30 frame pointer$ra 31 return address

Other registers are: $at = $1 which is reserved for the Assemblerand $k0, $k1 = $26, $27 which are reserved for the Operating System


The memory

• The memory is a long array• The address is the index to the array• Byte addressing - The index is in bytes

• The maximal size of memory available is

230 words = 232 bytes 0

1

2

3

4

5

6...

8 bits of data

8 bits of data

8 bits of data

8 bits of data

8 bits of data

8 bits of data

8 bits of data


Accessing the memory

• Load and Store instructions only• in LW we load a word, but the address we access is

given in bytes

lw $s1,100($s2) # $s1=Memory[$s2+100]

sw $s1,100($s2) # Memory[$s2+100]=$s1

offset = the location in the array base register = pointer to the array

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Accessing bytes

• There are also lb (load byte) and

sb (store byte) instructions

• These are useful for handling “char”s since ASCII characters are stored in bytes

ASCII: American Standard Code For information Interchange

• Another code, Unicode, store characters in two bytes each


ASCII


Accessing memory

• An example:

A is an array of words

The address of A is in $3

h is in $2

C code: A[2] = h + A[2];

MIPS code: lw $t0, 8($s3) # $t0=$s3[8] add $t0, $s2,$t0 # $t0=$s2+$t0 sw $t0, 8($s3) # $s3[8]=$t0

0

4

8

12

16

32 bits of dataA

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A[0]

A[1]

INTEL = Little endian

A[2]

MIPS = Big endian

013

0 1 2 3

2


• All instructions have the same size, 32 bits • In Intel’s 8086 the size changes from 1 to 17 bytes• An R-type instruction:

add $s1,$s2,$s3 # $s1=$17,$s2=$18, $s3=$19

Format of R-type instruction: 0 18 19 17 0 32

31 0

000000 10010 10011 10001 00000 100000

op rs rt rd shamt funct

op - opecode rs - register source

rt- register source no 2 rd - register destination

funct - function shamt - shift amount

Machine language

6 5 5 5 5 6

20H11H13H12H0H 0H


• Minimize the types. Use similar instructions.

• Example: lw $s1, 32($s2) # $s1 =$17, $s2=18

35 18 17 32

op rs rt 16 bit number

3rd design rule: A good design requires some compromises

op rs rt rd shamt funct

op rs rt 16 bit address

op 26 bit address

R

I

J

6 5 5 5 5 6

6 5 5 5 5 6

The compromise here is using different dormats to different types of instructions. We could have increased the number of bits per instruction so we have enough bits for all fields of all types of instructions. However, many of them would have been unused most of the time. So we use 3 types . (A different size conflicts with simplicuty).


• The program (code) is stored in memory just like data

Performing the program

• The Program Counter (PC) - a special register keeping the address of the next instruction

• We read the code word from the memory (using the PC as the pointer)

• We then increment the PC

Processor Memory

memory for data, programs, compilers, editors, etc.

The program in memory


• Jump - is an absolute jump without any conditionsj label

• Branch - a relative conditioned jump

bne $1,$2,label # $1!=$2 go to label

• An example:

if (i!=j) beq $s4, $s5, Lab1 h=i+j; add $s3, $s4, $s5else j Lab2 h=i-j; Lab1: sub $s3, $s4, $s5

Lab2: ...

Branch vs Jump

($s3=h $s4 =i $s5=j )


• Instructions:

bne $t4,$t5,Label Next instruction is at Label if $t4!= $t5

beq $t4,$t5,Label Next instruction is at Label if $t4 = $t5

j Label Next instruction is at Label

• Formats:

we see that

branch - is a relative jump in a range of 2^16 words

We assume that most of the jumps are local = branches

• Beq $s1,$s2,25 # if ($s1 ==$s2) go to PC +4 +25*4

op rs rt 16 bit address

op 26 bit address

I

J

Addresses in Branches and Jumps


Actual coding example

Loop: lw $8, save($19) # $8=save[i]

bne $8, $21,Exit #Goto Exit if save[i]<> k

add $19,$19,$20 # i:=i+j

j Loop # Goto Loop

Exit:

SAVE - 1000

1000 8 19 35 80,000

2 21 8 5 80,004

32 0 19 20 19 0 80,008

20,000 2 80,012

80,016


An important rule:

In MIPS Assembly instructions we write

code addresses in words

data addresses in bytes

when a MIPS CPU accesses memory it always gives the address in bytes

We cannot do:

bne $8,$21,far_adrs

since there are only 16 bits of offset in branch instructions.But we can do:

beq $8,$21,nxt j far_adrsnxt:

About long jumps:

This is a service given by the Assembler


slt $t0, $s1, $s2 means if $s1 < $s2 then $t0 = 1 else $t0 = 0

• We can use slt to “build” a blt instruction

blt $s0,$s1,Less

slt $at,$s0,$s1 # $t0 gets 1 if $s0<$s1

bne $at,$zero,Less # go to Less if $t0 != 0

• blt is a psedoinstruction

• Assembler uses $at (= $1) for pseudoinstructions

How do we code a Branch-if-less-than?

Also: move $t1,$s4 stands for: add $t1,$s4,$zero


Design rule 4: Build the common case faster

• Most arithmetic operation use small constants

• Large constants

addi $29, $29, 4

1010101010101010

0000000000000000 1010101010101010

1010101010101010 1010101010101010

0000000000000000 lui $t0,1010101010101010

ori $t0,$t0,1010101010101010

So we have many instructions with 16 bit constants: addi, slti, andi, ori, xori

These will be take longer to load, but only one more instruction is added


MIPS Assembly language summaryMIPS operands

Name Example Comments$s0-$s7, $t0-$t9, $zero, Fast locations for data. In MIPS, data must be in registers to perform

32 registers $a0-$a3, $v0-$v1, $gp, arithmetic. MIPS register $zero always equals 0. Register $at is $fp, $sp, $ra, $at reserved for the assembler to handle large constants.

Memory[0], Accessed only by data transfer instructions. MIPS uses byte addresses, so

230

memory Memory[4], ..., sequential words differ by 4. Memory holds data structures, such as arrays,

words Memory[4294967292] and spilled registers, such as those saved on procedure calls.MIPS assembly language

Category Instruction Example Meaning Commentsadd add $s1, $s2, $s3 $s1 = $s2 + $s3 Three operands; data in registers

Arithmetic subtract sub $s1, $s2, $s3 $s1 = $s2 - $s3 Three operands; data in registers

add immediate addi $s1, $s2, 100 $s1 = $s2 + 100 Used to add constants

load w ord lw $s1, 100($s2) $s1 = Memory[$s2 + 100]Word from memory to register

store w ord sw $s1, 100($s2) Memory[$s2 + 100] = $s1 Word from register to memory

Data transfer load byte lb $s1, 100($s2) $s1 = Memory[$s2 + 100]Byte from memory to register

store byte sb $s1, 100($s2) Memory[$s2 + 100] = $s1 Byte from register to memoryload upper immediate

lui $s1, 100 $s1 = 100 * 216 Loads constant in upper 16 bits

branch on equal beq $s1, $s2, 25 if ($s1 == $s2) go to PC + 4 + 100

Equal test; PC-relative branch

Conditional

branch on not equal bne $s1, $s2, 25 if ($s1 != $s2) go to PC + 4 + 100

Not equal test; PC-relative

branch set on less than slt $s1, $s2, $s3 if ($s2 < $s3) $s1 = 1; else $s1 = 0

Compare less than; for beq, bne

set less than immediate

slti $s1, $s2, 100 if ($s2 < 100) $s1 = 1; else $s1 = 0

Compare less than constant

jump j 2500 go to 10000 Jump to target address

Uncondi- jump register jr $ra go to $ra For sw itch, procedure return

tional jump jump and link jal 2500 $ra = PC + 4; go to 10000 For procedure call


RISC v. CISC

I - number of instructions in program

T - time of the clock cycle

CPI - number of clock cycles per instruction

RISC:

I T CPI

CISC:

I T CPI


An exercise

A C code and two MIPS Assembly translations are given below:

while (save[i]!=k) do

i=i+j ;

save:array [ 0..100] of word

k is kept in $21. $19 has i and $20 has j.

• First translation:

Loop: muli $9,$19,4 # Temporary reg $9:=i*4

lw $8,save($9) # Temporary reg $8:=save[i]

bne $8,$21,Exit # Goto Exit if save[i]<>k

add $19,$19,$20 # i:=i+j

j Loop # Goto Loop

Exit:


exercise (cont.)

• 2nd translation

muli $9,$19,4 # Temporary reg $9:=i*4


bne $8,$21,Exit # Goto Exit if save[i]<>k

Loop: add $19,$19,$20 # i:=i+j

muli $9,$19,4 # Temporary reg $9:=i*4


beq $8,$21,Loop # Goto Loop if save[i]=k

Exit:

• Assuming the loop is performed 10 times, what is the number of instructions performed in the two translations?


Compiler (Assembler) and Linker

A.asm

B.asm compiler

compilerA.obj

B.obj

linker

C.lib

(c.obj)

P.exeloader

Memory

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Linker operation

B.asm

compiler

compiler

A.obj

B.obj

s: .word 3,4

j k

lw $1,s ($2)

k: add $1,$2,$3

m: .word 2

sw 7, m($3)

A.asm

3 4

j 2

lw $1,0($2)

add $1,$2,$3

2

sw 7,0($3)2

3

4

sw 7,0($3)

j 3

lw $1,4($2)

add $1,$2,$3

linker

P.exe

Pages 156-160


The structure of an object file

• Object file header• text segment• data segment• relocation information• symbol table• debugging information

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1 Copyright 1998 Morgan Kaufmann Publishers, Inc. All rights reserved. MIPS Assembly language In...

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Transcript of 1 Copyright 1998 Morgan Kaufmann Publishers, Inc. All rights reserved. MIPS Assembly language In...