Answered! Consider the following fragment of ac program int [101, si int *pi for (p 131 p p++) Here is a buggy translation in…

int k[10], s;

int *p;

s=10;

for(p=&k[3]; *p!=0; p++)

s=s+p;

MIPS Code

or $16,$0,$0

lw $19,k+12

loop:

bne $8, finish

add $16,$19,$16

addi 19, 1

j loop

finish:

or $16,$0,$0

The syntax for the instruction ‘or’ is as follows

Or $d,$s,$t

$d=$s|$t

$16=$0|$0

$0 is a special purpose register so it contains value 0

$16=0|0

$16=0

It’s been told s is in $16

s=0

lw $19,k+12

The syntax for the instruction ‘lw’(load word) is as follows

lw $t, offset($s)

$t = MEM[$s + offset]

The register $t will contain the value present in the memory address given by offset +$s

The given instruction lw $19, k+12 is syntactically wrong.

bne $8, finish

The syntax for the instruction ‘bne’(branch on not equal) is as follows

bne $s,$t, offset

if($s!=$t) go to PC+4+offset

The given instruction bne $8, finish is syntactically wrong.

add $16,$19,$16

The syntax for the instruction ‘add’(add with overflow) is as follows

add $d, $s, $t

$d=$s+$t

add $16,$19,$16

$16=$19+$16

s is in $16 and p is in $19

s=p+s

from the first line(or $16,$0,$0) we know that s=0

0=p+0

0=p

addi $19, 1

The syntax for the instruction ‘addi’(add immediate) is as follows

addi $t, $s, imm

$t = $s + imm

The given instruction addi $19, 1 is syntactically wrong.

j loop

The syntax for the instruction ‘j’(jump to the calculated address) is as follows

j target

this is an unconditional jump statement