Convert C into MIPS

I wanted to convert this C code into MIPS.

C Code:

f = A[B[h-g]]

We assume that h > g & B[h-g] > 0. h,g,f are integers.

Also assume that f is assigned to register $s0, g to $s1, h to $s2.

Base addresses of A -> $s6 and B -> $s7

Here is my attempt:

sub $t0, $s2, $s1
mult $t0, $t0, 4
lw $t0, $t0($s7)
mult $t0, $t0, 4
sw $s0, $t0($s6)

Why do you post in a C forum if you want to write MIPS assembly?

oh...then where should I post this??

I would compile with the -S option (or whatever it is for your compiler) to see what the compiler generates. at least as an example.

It should probably be posted in the Tech Board. If a mod wants to move it...

You're almost there. I think that last line ought to be a lw as well. The way you have it written, you're storing $s0 in $t0($s6), i.e. reversing the assignment, putting f into A[B[h-g]]. You want to load A[B[h-g]] into $s0 (f). Other than that, it looks right, but it's been a while since I had anything to do with MIPS, so I wouldn't notice more subtle problems.

@anduril462 Yes it should be lw. Thanks a lot guys!

The corrected code is shown below:

Code:

---

sub $t0, $s2, $s1
mult $t0, $t0, 4
lw $t0, $t0($s7)
mult $t0, $t0, 4
lw $s0, $t0($s6)


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Just my 2c, but you might shave off some cycles by replacing the mult with sll:

Code:

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sub  $t0,$s2,$s1
sll  $t0,$t0,2
lw  $t0,$t0($s7)
sll  $t0,$t0,2
lw  $s0,$t0($s6)

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