What will be the output of the program?

#include<stdio.h>
void fun(int);
typedef int (*pf) (int, int);
int proc(pf, int, int);

int main()
{
    int a=3;
    fun(a);
    return 0;
}
void fun(int n)
{
    if(n > 0)
    {
        fun(--n);
        printf("%d,", n);
        fun(--n);
    }
}

Step 1: int no=5; The variable no is declared as integer type and initialized to 5.

Step 2: reverse(no); becomes reverse(5); It calls the function reverse() with '5' as parameter.

The function reverse accept an integer number 5 and it returns '0'(zero) if(5 == 0) if the given number is '0'(zero) or else printf("%d,", no); it prints that number 5 and calls the function reverse(5);.

The function runs infinetely because the there is a post-decrement operator is used. It will not decrease the value of 'n' before calling the reverse() function. So, it calls reverse(5) infinitely.

Note: If we use pre-decrement operator like reverse(--n), then the output will be 5, 4, 3, 2, 1. Because before calling the function, it decrements the value of 'n'.

Step 1: int fun(int); Here we declare the prototype of the function fun().

Step 2: int i = fun(10); The variable i is declared as an integer type and the result of the fun(10) will be stored in the variable i.

Step 3: int fun(int i){ return (i++); } Inside the fun() we are returning a value return(i++). It returns 10. because i++ is the post-increement operator.

Step 4: Then the control back to the main function and the value 10 is assigned to variable i.

Step 5: printf("%d\n", --i); Here --i denoted pre-increement. Hence it prints the value 9.

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.

Turbo C (Windows): The return value of the function is taken from the Accumulator _AX=1990.

But it may not work as expected in GCC compiler (Linux).

There is an error in this line i>=45 ? return(*p): return(*q);. We cannot use return keyword in the terenary operators.

Step 1: int check(int); This prototype tells the compiler that the function check() accepts one integer parameter and returns an integer value.

Step 2: int l=45, c; The variable i and c are declared as an integer type and i is initialized to 45.

The function check(i) return 100 if the given value of variable i is >=(greater than or equal to) 45, else it will return 10.

Step 3: c = check(i); becomes c = check(45); The function check() return 100 and it get stored in the variable c.(c = 100)

Step 4: printf("%d\n", c); It prints the value of variable c.

Hence the output of the program is '100'.

Step 1: int i=0; The variable i is declared as in integer type and initialized to '0'(zero).

Step 2: i++; Here variable i is increemented by 1. Hence i becomes '1'(one).

Step 3: if(i<=5) becomes if(1 <=5). Hence the if condition is satisfied and it enter into if block statements.

Step 4: printf("CuriousTab"); It prints "CuriousTab".

Step 5: exit(1); This exit statement terminates the program execution.

Hence the output is "CuriousTab".

Any function including main() can be called recursively.

True, A function cannot return more than one value at a time. because after returning a value the control is given back to calling function.