4. Translate the following MIPS into C. (1) Assume that the variables a, b, i, a

Question

4. Translate the following MIPS into C.

(1) Assume that the variables a, b, i, and j are stored in registers $s0, $s1, $t0, and $t1 respectively. Assume that registers $s2 and $s3 hold the base registers for arrays A and B.

          addi $t1, $zer0, 1000                  

loop: slt $t0, $s1, $t1                           

          bne $t0, $zer0, EXIT                   

          sll $t2, $t1, 2                               

          add $t2, $t2, $s2                         

          lw $t2, 0($t2)                               

          add $s1, $s1, $t2                        

          addi $t1, $t1, -1                           

          j loop

EXIT:

(2)

A:       addi $sp, $sp, -4

          sw $s0, 0($sp)

          add $s0, $a0, $a1

          mul $s0, $s0, $s0

          addi $s0, $s0, 1234

          add $v0, $s0, $zer0

          lw $s0, 0($sp)

          addi $sp, $sp, 4

          jr $ra

Explanation / Answer

1)
int x=0;    //$t2=x
int * addr_A = array_A;   //pointer addr_A have the base address of array A
j=1000;
while(1){
    if(b < j){
        i=1;
    }
    if(i != 0){
        break
    }
    x= j << 2;
    addr_A = addr_A + x;
    b=b + *(addr_A);
    j--;
}

2)

NOTE:- As variable notation not given;I am writing in here
$a0,$a1 = a,b (which are function arguement)

int A(int a,int b){
    int x=a+b;
    x= x * x;
    x= x + 1234
    return x;
}

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