Q1:

What will be the output of the program?

#include<stdio.h>

int addmult(int ii, int jj)
{
    int kk, ll;
    kk = ii + jj;
    ll = ii * jj;
    return (kk, ll);
}

int main()
{
    int i=3, j=4, k, l;
    k = addmult(i, j);
    l = addmult(i, j);
    printf("%d, %d\n", k, l);
    return 0;
}

Step 1: int i=3, j=4, k, l; The variables i, j, k, l are declared as an integer type and variable i, j are initialized to 3, 4 respectively. The function addmult(i, j); accept 2 integer parameters. Step 2: k = addmult(i, j); becomes k = addmult(3, 4) In the function addmult(). The variable kk, ll are declared as an integer type int kk, ll; kk = ii + jj; becomes kk = 3 + 4 Now the kk value is '7'. ll = ii * jj; becomes ll = 3 * 4 Now the ll value is '12'. return (kk, ll); It returns the value of variable ll only. The value 12 is stored in variable 'k'. Step 3: l = addmult(i, j); becomes l = addmult(3, 4) kk = ii + jj; becomes kk = 3 + 4 Now the kk value is '7'. ll = ii * jj; becomes ll = 3 * 4 Now the ll value is '12'. return (kk, ll); It returns the value of variable ll only. The value 12 is stored in variable 'l'. Step 4: printf("%d, %d\n", k, l); It prints the value of k and l Hence the output is "12, 12".