Add comments to the code and describe in one sentence what this code does. Assume that $a0 and $a1 are used for input and both initially contain the integers a and b, respectively. Assume that $v0 is used for the output. Also, convert this MIPS code to C

add $t0, $zero, $zero

loop: beq $a1, $zero, finish

add $t0, $t0, $a0

addi $a1, $a1, - 1

j loop

finish: addi $t0, $t0, 100

add $v0, $t0, $zero

Code is given below:

Explanation:

It does a0*a1+100

int func (int a0, int a1, int a2, int a3) / /In mips $a0 to $a4 are arguments to function

{

const int zero = 0; / / $zero is register whose value is always 0

int t0 = zero; //t0 is temporary register

while (a1!=0) / /beq is like a1! = 0 go out of loop

{

t0 = t0 + a0; //add a0 to "to" a1 times which is like a0*a1

a1 = a1 - 1;

}

t0 = t0 + 100; //a0*a1+100

return t0; / / return this value

}

int main ()

{

int a0, a1;

printf ("Enter a0 and a1/n");

scanf ("%d %d",&a0,&a1);

printf ("%d*%d+100=%d/n", func (a0, a1,0,0));

}