For the following MIPS code write the C equivalent: The following code fragment processes two arrays and produces an important value in register $v0. Assume that each array consists of 2500 words indexed 0 through 2499, that the base addresses of the arrays are stored in $a0 and $a1 respectively, and their sizes (2500) are stored in $a2 and $a3 , respectively. Add comments to the code and describe in one sentence what this code does. Specifically, what will be returned in $v0 ? Also, convert this MIPS code to C (you may find it easier to understand if you do this first!).

//get size of array one

int size1 =2500; // sll $a2,$a2,2

int array1[size1];

//get size of array two

int size2 = 2500; // sll $a3,$a3,2

int array2[size2];

int v0=0 ; // $v0,$zero,$zero

int t0 = 0; //$t0,$zero,$zero

do{ //outer:

//add $t4,$a0,$t0

//lw $t4,0($t4)

int t4 = array1[t0];

int t1 = 0; //$t1,$zer0,$zer0

do{

//add $t3,$a1,$t1

//lw $t3,0($t3)

int t3 = array2[t1];

if(t3==t4){ //bne $t3,$t4,skip

v0 = v0+1; //addi $v0,$v0,1

}

//skip:

t1 = t1+1; //addi $t1,$t1,4

}while(t1!=size2);

t0 = t0+1; //addi $to,$to,4

}while(t0!=size1);

return 0;

Given code loops through the two array. It will count the number of duplicates found in both array.