MIPS Assembly to C

Question

So I'm trying to translate this MIPS assembly code into C. I'm confused on a certain part of what's going on. Here is the MIPS assembly code: Assume we have variables f, g, h, i, j stored in $s0, $s1, $s2, $s3 and $s4, respectively. Assume the base addresses of arrays A and B are at $s6 and $s7 respectively and they contain 4 byte words. I have inserted comments to show I understand most of this.

sll  $t0, $s0, 2    # $t0 = f * 4
add  $t0, $s6, $t0  # $t0 = &A[f]
sll  $t1, $s1, 2    # $t1 = g * 4
add  $t1, $s7, $t1  # $t1 = &B[g]
lw   $s0, 0($t0)    # f = A[f]
addi $t2, $t0, 4    <-- Here's where I am confused. Since $t0 contains the address of A[f], what does adding 4 do to that?

lw   $t0, 0($t2)    
add  $t0, $t0, $s0
sw   $t0, 0($t1)

Jakub Świerk · Accepted Answer · 2013-09-17 23:29:28Z

3

It's look like to calculate pointer to next element of A array, one line later you load data from this address. Assuming in C:

Word4Byte A[],B[];
B[g] = A[f] + A[f+1];

answered Sep 17, 2013 at 23:29

Jakub Świerk

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