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));

}

answer; (& & );

b. xor is the correct answer