Problem in translating from MIPS to C

I was trying to solve this homework assignment but was unable to come up with a solution. Below is the problem,

Translate the following MIPS code into a high-level language program. Assume that $t0, $t1, and $t2 contain base addresses of arrays A, B and C respectively.

add $t4, $zero, $zero
Loop:
add $t5, $t4, $t1
lw $t6, 0($t5)
add $t5, $t4, $t2
lw $t7, 0($t5)
add $t6, $t6, $t7
add $t5, $t4, $t0
sw $t6, 0($t5)
addi $t4, $t4, 4
slti $t5, $t4, 256
bne $t5, $zero, Loop

Also find the decimal value in the offset field of bne

instruction.

Here is what I have tried, but I yet don't find the position of 256.

int *temp4 = 0;
while(1)
{
    *int temp5 = temp4 +B[0];
    a:
        *int temp6 = temp5;
        temp5 = C[0] + temp4;
        *temp7 = temp5;
        temp6 = temp6 + temp7;
        temp5 = temp4 + A[0];
        temp6 = temp5;
        temp4 += 4;
  开发者_如何转开发      if(temp5 < temp4)
            goto __;
        if(temp5 != 0)
            goto a;
}

I think that you are overthinking things.

What that code does is something like this

for (int i =0 ; i< 64; i++){

   A[i] = B[i] + C[i];

}

Not going to explain why, since this looks a lot like is homework.

Here is what I have tried, but I yet don't find the position of 256.

You are mixing things up. See that i at the end of slti ? That means that it uses an inmediate operand, in this case 256. So what the slti $t5, $t4, 256 instruction is doing, is setting a 1 in register $t5 if the contents of $t4 are lower than 256. Otherwise, $t5 gets a 0.

Consequently, the loop will take 256/4 iterations, since the bne will fallthrough (i.e not jump) only when the contents of $t4 are greater than 256.