Q1:

What will be the output of the program?

#include<stdio.h>
void fun(int*, int*);
int main()
{
    int i=5, j=2;
    fun(&i, &j);
    printf('%d, %d', i, j);
    return 0;
}
void fun(int *i, int *j)
{
    *i = *i**i;
    *j = *j**j;
}

Step 1: int i=5, j=2; Here variable i and j are declared as an integer type and initialized to 5 and 2 respectively. Step 2: fun(&i, &j); Here the function fun() is called with two parameters &i and &j (The & denotes call by reference. So the address of the variable i and j are passed. ) Step 3: void fun(int *i, int *j) This function is called by reference, so we have to use * before the parameters. Step 4: *i = *i**i; Here *i denotes the value of the variable i. We are multiplying 5*5 and storing the result 25 in same variable i. Step 5: *j = *j**j; Here *j denotes the value of the variable j. We are multiplying 2*2 and storing the result 4 in same variable j. Step 6: Then the function void fun(int *i, int *j) return back the control back to main() function. Step 7: printf('%d, %d', i, j); It prints the value of variable i and j. Hence the output is 25, 4.