Q1: What will be the output of the program?

#include<stdio.h>

int fun(int i)
{
    i++;
    return i;
}

int main()
{
    int fun(int);
    int i=3;
    fun(i=fun(fun(i)));
    printf("%d\n", i);
    return 0;
}

ANS:A - 5

Step 1: int fun(int); This is prototype of function fun(). It tells the compiler that the function fun() accept one integer parameter and returns an integer value. Step 2: int i=3; The variable i is declared as an integer type and initialized to value 3. Step 3: fun(i=fun(fun(i)));. The function fun(i) increements the value of i by 1(one) and return it. Lets go step by step, => fun(i) becomes fun(3) is called and it returns 4. => i = fun(fun(i)) becomes i = fun(4) is called and it returns 5 and stored in variable i.(i=5) => fun(i=fun(fun(i))); becomes fun(5); is called and it return 6 and nowhere the return value is stored. Step 4: printf("%d\n", i); It prints the value of variable i.(5) Hence the output is '5'.