What will be the output of the program?

#include<stdio.h>

int main()
{
    int fun(int);
    int i = fun(10);
    printf("%d\n", --i);
    return 0;
}
int fun(int i)
{
   return (i++);
}

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).

Step 1: int k=35; The variable k is declared as an integer type and initialized to 35.

Step 2: k = func1(k=func1(k=func1(k))); The func1(k) increement the value of k by 1 and return it. Here the func1(k) is called 3 times. Hence it increements value of k = 35 to 38. The result is stored in the variable k = 38.

Step 3: printf("k=%d\n", k); It prints the value of variable k "38".

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'.

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'.

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'.