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));

}

is this supposed to be a question ?

solution:

this situation is called an ip conflict.

data from a server is sent in small bits called packets. in an ip conflict, some packets will end up in one computer while some other packets will end up in another computer.

so…overall none of the computers get the data as the original message or picture being sent by the server cannot be retrieved with only half the packets. so it will show up as an error.

thus the final answer is d.