C to MIPS Conversion Practice Set 2

Full MIPS solutions are provided below so you can blank out fields as needed.

Assumptions:

• Calling convention: arguments in $a0, $a1, $a2 and the return value should be placed in $v0
• 32-bit MIPS

1) Function f1

C code:

/* p: $a0, t: $a1, v: $a2 */
int f1(short *p, unsigned int t, short v)
{
    int x = v + *p;
    t = t - 2;
    *p = x >> t;
    return x;
}

MIPS solution:

f1:
    l__       $t0, 0($a0)
    add       ___, $a2, $t0
    addi      $t1, $a1, -2
    ____v     $t2, ___, $t1
    ___       $t2, _____
    jr        $ra

2) Function f2

C code:

/* arr: $a0, k: $a1, j: $a2*/
void f2(int *arr, int k, int j)
{
    if (k < j) {
        arr[0] = arr[k];
    }
    else {
        arr[1]++;
    }
}

MIPS solution:

f2:
    ____      $t0, $a1, $a2
    b____     $t0, $zero, L1
    sll       $t1, $a1, ____
    add       $t1, $a0, $t1
    lw        $t2, ____
    sw        $t2, ____
    b         L2
L1:
    ____      $t3, ____
    addi      $t3, $t3, 1
    sw        ____, ____
L2:
    jr        $ra

f1 Solutions

f1:
    lh        $t0, 0($a0)
    add       $v0, $a2, $t0
    addi      $t1, $a1, -2
    srav      $t2, $v0, $t1
    sh        $t2, 0($a0)
    jr        $ra

f2 Solutions

f2:
    slt       $t0, $a1, $a2
    beq       $t0, $zero, L1
    sll       $t1, $a1, 2
    add       $t1, $a0, $t1
    lw        $t2, 0($t1)
    sw        $t2, 0($a0)
    b         L2
L1:
    lw        $t3, 4($a0)
    addi      $t3, $t3, 1
    sw        $t3, 4($a0)
L2:
    jr        $ra