What will be the output of the program in 16 bit platform (Turbo C under DOS)?

#include<stdio.h>

int main()
{
    int fun();
    int i;
    i = fun();
    printf("%d\n", i);
    return 0;
}
int fun()
{
    _AX = 1990;
}

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 i=1; The variable i is declared as an integer type and initialized to 1(one).

Step 2: if(!i) Here the !(NOT) operator reverts the i value 1 to 0. Hence the if(0) condition fails. So it goes to else part.

Step 3: else { i=0; In the else part variable i is assigned to value 0(zero).

Step 4: printf("C-Program"); It prints the "C-program".

Step 5: main(); Here we are calling the main() function.

After calling the function, the program repeats from step 1 to step 5 infinitely.

Hence it prints "C-Program" infinitely.

Step 1: int i; The variable i is declared as an integer type.

Step 1: int fun(); This prototype tells the compiler that the function fun() does not accept any arguments and it returns an integer value.

Step 1: while(i) The value of i is not initialized so this while condition is failed. So, it does not execute the while block.

Step 1: printf("Hello\n"); It prints "Hello".

Hence the output of the program is "Hello".

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.

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 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'.