c code
while (save[i]==k)
i+=1;
put i in $s3, k in $s5 and address of k in $s6
mips code
loop: sll $t1, $s3, 2
add $t1, $t1, $s6
lw $t0, 0($t1)
bne $t0, $s5, Exit
addi $s3, $s3, 1
j loop
Exit:
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Ahmed Butt
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To answer this question we need to see the declarations of `i` and `save`. – zwol Nov 19 '21 at 19:18
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Please provide enough code so others can better understand or reproduce the problem. – Community Nov 23 '21 at 17:49
1 Answers
2
Your save array/pointer must be of some 4-byte type (ints?). Therefore to load save[i] from memory, the index i needs to be translated to a byte offset within the array, and then added to the base address of that array. This is done by multiplying i by four:
sll $t1, $s3, 2
and then adding save:
add $t1, $t1, $s6
This doesn't look like an optimized build though. Usually the compiler can re-write this code to advance a temporary pointer in increments of four directly, thus avoiding two of the instructions in that loop.
Yakov Galka
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Indeed, the loop in the question is really inefficient, and could have used `lw` / `add`(pointer increment) / `bne loop_top` on a fake MIPS without branch-delay slots to at least hide some load latency, and avoid the unconditional branch. – Peter Cordes Nov 20 '21 at 00:51
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@PeterCordes I actually don't know MIPS instruction set, but that looked unoptimized even to my eye :) – Yakov Galka Nov 20 '21 at 02:25