Mips instruction set
Download
1 / 11

MIPS Instruction Set - PowerPoint PPT Presentation


  • 164 Views
  • Updated On :

MIPS Instruction Set. Conditional Expressions and Branching. Outline. Motivation for conditional expressions and loops How to add a sequence of numbers Conditional expressions (branching) Jumping Switching. Last time ...adding 3 numbers. What a mess this is...

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'MIPS Instruction Set' - jerrell


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
Mips instruction set

MIPS Instruction Set

Conditional Expressions and Branching


Outline
Outline

  • Motivation for conditional expressions and loops

  • How to add a sequence of numbers

  • Conditional expressions (branching)

  • Jumping

  • Switching


Last time adding 3 numbers
Last time ...adding 3 numbers...

  • What a mess this is...

    • lw $10,A($0) # $10 = a0 ($0 always contains 0)

    • li $4, 4 # $4 = 4

    • lw $3,A($4) # $3 = a1

    • add $10,$10,$3 # $10 = a0 + a1

    • li $4, 8 # $4 = 8

    • lw $3,A($4) # $3 = a2

    • add $10,$10,$3 # $10 = a0 + a1+ a2

    • li $4, 12 # $4 = 12

    • sw $10,A($4) # a3 gets $10

  • Suppose 100 numbers to be added?

  • Doesn’t generalise!


Adding a sequence of n numbers
Adding a sequence of n numbers

  • x[0], x[1], ..., x[n-1] sequence of n numbers.(n=100, say).

  • Consider (abstract representation)

    sum = 0 # initialise sum

    n = 100 # initialise boundary condition

    i = 0 # initialise counter

    LOOP: # label

    sum = sum + x[i] #overwrite sum

    i = i+1 #increment counter

    if (i != n) then go to LOOP


Adding the sequence in mips
Adding the Sequence in MIPS

move $5,$0 # puts 0 into $5 (sum)

addi $4,$0,100 # puts 100 into $4 (n)

move $11,$0 # puts 0 into $11 (i)

li $12,4 # puts 4 into $12

LOOP: mult $14,$11,$12 # $14 = i*4 (why?!)

lw $6,Xstart($14) # retrieves x[i]

add $5,$5,$6 # sum = sum + x[i]

addi $11,$11,1 # i = i+1

bne $11,$4, LOOP # if i < n go to LOOP

sw $5,Sum($0) #result put in memory


Conditional expressions
Conditional Expressions

  • bne $a,$b,LABEL

    • branch if not equal

      • if $a != $b then jump to LABEL

  • beq $a,$b,LABEL

    • branch if equal

      • if $a == $b jumpt to LABEL

  • slt $a,$b,$c

    • set if less than

      • if ($b < $c) then $a = 1 else $a = 0


Example introduces jump
Example (introduces jump)

  • Consider the following abstract code

    if (a < b) then c = 100

    else c = 500

  • In MIPS (suppose a,b,c are $5,$6,$7)

    slt $10,$5,$6 # $10 = 1 if $5<$6

    beq $10,$0,ELSE # if $10==0 goto else

    addi $7,$0,100 # $7 =100

    j CONTINUE # jump to continue

    ELSE : addi $7,$0,500 # $7 = 500

    CONTINUE: whatever here...


Choosing from several alternatives
Choosing from several alternatives

  • Abstract code

    if k=0 then a=20

    else if k=1 then a=10

    else if k=2 then a=11;

  • C/C++ representation (uses ‘switch’)

    switch(k){

    0 : a = 20; break;

    1 : a = 10; break;

    2 : a = 11;

    }


Using jump register jr a
Using jump register (jr $a)

  • jumps to instruction referenced in $a

  • suppose Label is memory location containing addresses L0, L1, L2. ($4=k*4)

    lw $10,Label($4) # $10 = Label[k]

    jr $10 # jump $10

    L0: addi $20,$0,20 # $20 gets 20

    j BREAK

    L1: addi $20,$0,10 # $20 gets 10

    j BREAK

    L2: addi $20,$0,11 # $20 gets 11

    BREAK: continue whatever here


Summary
Summary

  • bne $a,$b,LABEL #branch if not equal

  • beq $a,$b,LABEL #branch if equal

  • slt $a,$b,$c # select if less than

  • j LABEL #jump to the instruction referenced by LABEL

  • jr $a #jump to instruction referenced in register $a


Still potential for a mess
Still potential for a mess!...

  • Adding 100 numbers x[] is fine.

  • Suppose later we want to add 1011 numbers using sequence y[]?

    • Do we have to repeat the same set of instructions all over again, with small modifications?

    • No way!

  • Use procedures!!!