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HW #1 Problems & Industrial Strength Assembly CSCI 136

HW #1 Problems & Industrial Strength Assembly

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HW #1 Problems & Industrial Strength Assembly. CSCI 136. Reading the Question. “If there are no b’s in the string, then bfind should return a pointer to the null character at the end of the string” lbu $t2, 0($t3) beq $t2, $0, term … term: la $a0, ndone li $v0 ,4 - PowerPoint PPT Presentation

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Page 1: HW #1 Problems & Industrial Strength Assembly

HW #1 Problems &Industrial Strength Assembly

CSCI 136

Page 2: HW #1 Problems & Industrial Strength Assembly

Reading the Question“If there are no b’s in the string, then bfind

should return a pointer to the null character at the end of the string”

lbu $t2, 0($t3) beq $t2, $0, term …term: la $a0, ndone li $v0 ,4 syscallIts great that you tell me whats happening..

How about putting $t3 in to $v0?

Page 3: HW #1 Problems & Industrial Strength Assembly

Reading the Question“The bfind procedure should locate the first b

character in the string and return its address in register $v0”

bfind: lbu $t1,0($a0)

beqz $t1, Ret

beq $t1, ‘b’, Ret

addi $a0, $a0, 1

j bfind

Ret: sb $v0, ($t1)

Page 4: HW #1 Problems & Industrial Strength Assembly

Using the bfind procedure in bcount“You must use your bfind procedure in

Exercise 3.23 in your implementation of bcount.”

What does bfind return?

So how do we check if we’re at the end of the string?

jal bfindlb $t0,0($v0)beqz $t0, endOfString

Page 5: HW #1 Problems & Industrial Strength Assembly

RTFQ – Reading the QuestionTough parts of Software:

1. Specifying Requirements

2. Understanding the Specified Requirements

Why it matters at GWU: Grading Scripts

Besides.. It should be really frustrating to spend hours answering the wrong question…

Page 6: HW #1 Problems & Industrial Strength Assembly

RTFQ- Reading the QuestionTypical Errors:Didn’t write bcount as a procedure:

This requires saving $ra before the call to bfind, and restoring $ra after the call to bfind

main: jal bcount

bcount: jal bfindWhat happens to $ra when jal is called?

Page 7: HW #1 Problems & Industrial Strength Assembly

Answering the Question add $t0, $0, $0

bfind: add $t3, $t0, $a0

lbu $t2, 0 ($t3)

beq $t2, $0, done

addi $t2, $t2, -98

beq $t2, $0, done

addi $t0, $t0, 1

j bfind

done: addi $t0,1

add $v0,$t0,$0

jr $ra

Page 8: HW #1 Problems & Industrial Strength Assembly

Answering the Question li $t0,0

bfind: add $t3,$a0,$t0

lbu $t2, 0 ($t3)

beq $t2, $0, done

beq $t2, ‘b’, done

addi $t0, $t0, 4

j bfind

done: move $s2,$t0

jr $ra

Page 9: HW #1 Problems & Industrial Strength Assembly

Why do we program in Assembly?

Page 10: HW #1 Problems & Industrial Strength Assembly

Why do we program in Assembly?

What matters in real world:

SPEED, SPEED & SPEED

Other acceptable answer:Low Level Interface with hardware (mouse,

ethernet card, serial port, video card, etc.)Unfortunately we can’t simulate these

hardware interfaces with SPIM.

Page 11: HW #1 Problems & Industrial Strength Assembly

Naïve Approach to Style

“Good Style” == lots of comments

Even poor code well

documented is still

poor code.

Page 12: HW #1 Problems & Industrial Strength Assembly

Style• Accomplish task in straight forward manner

with minimum effort

• Unfortunately, “Good Style” comes from experience and experience comes from “Bad Style”

We don’t expect you to have wonderful style in the first couple assignments…

By the end of semester you should be writing simple straight forward code.

Page 13: HW #1 Problems & Industrial Strength Assembly

Minimum Effort

la $t0, $a0bfind: lbu $t1, 0 ($t0) beq $t1, Done beq $t1,’b’, Done addi $t0, 1 addi $a0, 1 j bfind

Done: move $v0,$a0

Page 14: HW #1 Problems & Industrial Strength Assembly

Minimum EffortRegisters are a Commodity. Do not waste

them!Why is minimizing the use of registers so

important?

What is the slowest type of instruction on our processor?

If we have registers available... We can use them.

This will allow your programs to run faster!

Page 15: HW #1 Problems & Industrial Strength Assembly

Minimum EffortRegisters are a Commodity. Do not waste

them!Why is minimizing the use of registers so important?

Modern processors have multiple execution units…

Processors can execute several different portions of your code at the same time...

Lack of registers prevents multiple execution…

Help the processor out by using minimal number of registers and having minimal register “overlap”.

This will allow your programs to run faster!

Page 16: HW #1 Problems & Industrial Strength Assembly

Do we need all these registers? li $s1,0x00

bfind:

lbu $t2,0($a0)

beq $t2,$s1,exit

beq $t2,’b’,exit

addi $a0,$a0,1

j bfind

exit:

Page 17: HW #1 Problems & Industrial Strength Assembly

Do we need all this code?bfind:

lbu $t2,0($a0)

addi $a0,$a0,1

beq $t2,$zero,exit

beq $t2,’b’,exit

j bfind

exit:

subu $a0,$a0,1

Page 18: HW #1 Problems & Industrial Strength Assembly

Minimum EffortKnow your architecture (load - store) and

what your pseudo instructions do:

Is there a difference difference between:

beq $t0, ‘b’, Done

and

addi $at,$zero,’b’

beq $t0,$at, Done

Page 19: HW #1 Problems & Industrial Strength Assembly

Do I need everything in my loop?bfind: lbu $t0, 0 ($a0)

beqz $t0, done

addi $t1,$zero,’b’

beq $t0, $t1, done

add $a0, $a0, 1

j bfind

done: mov $v0, $a0

jr $ra

Page 20: HW #1 Problems & Industrial Strength Assembly

Extra Branches & Border Case Failurebfind: li $t1, ‘b’ lb $t0,0($a0) bne $t0,$t1,loop move $v0,$a0quit: jr $raloop: addi $a0,$a0,1 lbu $t0,0($a0) beq $t0,$zero,quit bne $t0,$t1,loop move $v0,$a0 jr $ra

Page 21: HW #1 Problems & Industrial Strength Assembly

Extra Branches and Loops

Why are extra branches bad?

-Optimizing Branches

Why are extra loops bad?

Page 22: HW #1 Problems & Industrial Strength Assembly

Public and Private VariablesIn SPIM- When you call a function… and the function

returns… the only values from that function your procedure should access are:

$v0, $v1

Values saved by the function on the stackLocations in Memory

DO NOT access the internal variables (registers) of another function! This “breaks” the whole “function” paradigm. It isn’t a “black box” if you go digging around…

Page 23: HW #1 Problems & Industrial Strength Assembly

Public and Private Variables

FUNCTION

GWU

Page 24: HW #1 Problems & Industrial Strength Assembly

Breaking the Function Paradigmaddi $sp, $sp, -4

sw $ra,0($sp)

jal bfind

lw $ra,0($sp)

addi $sp,$sp,4

quit: bne $t0, $zero, adder

Who set $t0? bfind did. But the only return value was supposed to be $v0?

What are the problems with this?

Page 25: HW #1 Problems & Industrial Strength Assembly

Breaking the Function Paradigm

I should be able to plug in ANY bfind in to bcount and make it work.

Don’t fight the framework!

Page 26: HW #1 Problems & Industrial Strength Assembly

Real World Disclaimer…Register assignments aren’t as structured…(no $t0-$t9, $s0-$s7..just registers $1-$31)

Saving values on stack or in memory is slow.

More registers can be used to return values.

Function writers and function callers typically have specific agreements about what registers to use for passing and returning values.

Page 27: HW #1 Problems & Industrial Strength Assembly

Proper Instructionsjr $1 is jump to a value given by the register $1

(Where do you typically see jr $ra ?)

j is jump to a label

Use addi to add a constant to a register

Use add to add registers together

Watch the ordering of registers in instruction…

sw $sp, $ra What does this do?

Page 28: HW #1 Problems & Industrial Strength Assembly

Good Implementation of Bfind addi $t1,$zero,’b’bfind: lbu $t0, 0 ($a0) beqz $t0, done beq $t0, $t1, done add $a0, $a0, 1 j bfind

done: move $v0, $a0 jr $ra8 lines of assembly. No extra branches,

loops, variables. Common case is fast.

Page 29: HW #1 Problems & Industrial Strength Assembly

Lab Assignment: Endian Fixer• TCP/IP message headers are in Big Endian• Your PC is a Little Endian machine• Write function that takes word value in $a0 and

returns a Endian reversed value in $v0• Goals:

– Learn how to use masks

li $t0, 0x000000ff

andi $t1, $a0, $t0

ori $v0, $t0, $t1

Page 30: HW #1 Problems & Industrial Strength Assembly

endianFixer: #1234rol $a0,$a0,8 #2341lui $t1, 0x00ffori $t1, $t1, 0x00ffand $v0,$t1,$a0 #0301

sll $t1,$t1,8 #$t1=ff00ff00rol $a0,$a0,16 #4123and $a0,$t1,$a0 #4020or $v0,$v0,$a0 #4321

jr $ra Why don’t I ever zeroize $v0?Is li $t1,0x00ff00ff ok?

Page 31: HW #1 Problems & Industrial Strength Assembly

Adder Review• How many delays does a 4 bit ripple carry

adder take to add two 4 bit numbers? Why?

Page 32: HW #1 Problems & Industrial Strength Assembly

Adder ReviewWhat are the equations for a 4 bit carry look

ahead adder? Do you understand this?

Page 33: HW #1 Problems & Industrial Strength Assembly

From when a and b’s show up… how long does it take until sum appears on the output lines?

How long for carry generation + how long for sum?

Page 34: HW #1 Problems & Industrial Strength Assembly

8 bit adderWe need an 8 bit adder…how can we do this?

Page 35: HW #1 Problems & Industrial Strength Assembly

8 bit adder We could have a pure ripple carry…

How long would this take to calculate

the sums?

Page 36: HW #1 Problems & Industrial Strength Assembly

8 bit adder We could use ripple carry to connect

two 4 bit carry look ahead groups…

How long would this take to calculate

the carries? The sums?

Page 37: HW #1 Problems & Industrial Strength Assembly

2nd layer of Carry LookaheadLets add another layer of Carry Lookahead for

our 8 bit adder…

What are the equations for big Ps?

What are the equations for big Gs?

What are the equations for the big Cs?

Page 38: HW #1 Problems & Industrial Strength Assembly

2nd layer of Carry LookaheadHow long would it take to calculate the carries?

How long would it take to calculate the sums?