CODEWITHSUNDEEP
cmips
user1631224
user1631224

Reputation: 409

Is this MIPS assembly code converted from C code correct?

I'm having trouble converting a C code, that contains a function call to another function, into MIPS.

Here is the info:

The function declaration for go is int now(int i, j). The C code for the function now is:

int now(int i, int j, int k, int l){
    return go( go(i, j), k+l);
}

Here is my MIPS code so far:

now:
    addi      $sp,   $sp,      -8       # adjust the stack to store 2 items
    sw        $ra,   4($sp)             # save return address of func(i, j)
    sw        $t0,   0($sp)             # save register for use afterwards
    move      $a0,   $a0
    move      $a1,   $a1
    jal       go
    add       $t0,   $a2,      $a3      # register $t0 = k + l
    move      $a1,   $t0
    jal       go
    lw        $t0,   0($sp)
    lw        $ra,   4($sp)
    addi      $sp,   $sp,      8
    jr        $ra

I'm having trouble with go(i, j) being passed as a parameter for the function go

Any inputs would be appreciated.

Thanks!

Upvotes: 2

Views: 149

Answers (1)

Ross Ridge
Ross Ridge

Reputation: 39551

This question probably doesn't belong on this site, but here's my comments.

    sw        $t0,   0($sp)             # save register for use afterwards

You don't need to save $t0, as it's assumed to be clobbered by functions.

    move      $a0,   $a0
    move      $a1,   $a1

These two moves do nothing so remove them.

    jal       go

The call may clobber a2 and a3.

    add       $t0,   $a2,      $a3      # register $t0 = k + l

Move this statement before the call and change the register to $s0 and save that register instead of saving $t0. If you do that it doesn't matter if a2 or a3 get clobbered because you don't need their values anymore.

    move      $a1,   $t0
    jal       go

You didn't give anything as the first argument in the call. Move the return value of the previous call from $v0 to $a0. Load $a1 with the value stored in $s0 as recommended above.

    lw        $t0,   0($sp)
    lw        $ra,   4($sp)
    addi      $sp,   $sp,      4
    jr        $ra

You saved 2 registers for a total of 8 byte but you're only adding 4 to $sp.

Upvotes: 2

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