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;
}

A const variable has to be initialized when it is declared. later assigning the value to the const variable will result in an error "Cannot modify the const object".

Hence Option B is correct

Step 1: A macro named MAX is defined with value 128

Step 2: const int max=128; The constant variable max is declared as an integer data type and it is initialized with value 128.

Step 3: char array[max]; This statement reports an error "constant expression required". Because, we cannot use variable to define the size of array.

To avoid this error, we have to declare the size of an array as static. Eg. char array[10]; or use macro char array[MAX];

Note: The above program will print A A as output in Unix platform.

This program will show an error "Cannot modify a const object".

Step 1: const int i=0; The constant variable 'i' is declared as an integer and initialized with value of '0'(zero).

Step 2: printf("%d\n", i++); Here the variable 'i' is increemented by 1(one). This will create an error "Cannot modify a const object".

Because, we cannot modify a const variable.

Step 1: char mybuf[] = "India"; The variable mybuff is declared as an array of characters and initialized with string "India".

Step 2: char yourbuf[] = "CURIOUSTAB"; The variable yourbuf is declared as an array of characters and initialized with string "CURIOUSTAB".

Step 3: char *const ptr = mybuf; Here, ptr is a constant pointer, which points at a char.

The value at which ptr it points is not a constant; it will not be an error to modify the pointed character; There will be an error only to modify the pointer itself.

Step 4: *ptr = 'a'; The value of ptr is assigned to 'a'.

Step 5: ptr = yourbuf; Here, we are changing the pointer itself, this will result in the error "cannot modify a const object".

Step 1: char mybuf[] = "India"; The variable mybuff is declared as an array of characters and initialized with string "India".

Step 2: char yourbuf[] = "CURIOUSTAB"; The variable yourbuf is declared as an array of characters and initialized with string "CURIOUSTAB".

Step 3: char const *ptr = mybuf; Here, ptr is a constant pointer, which points at a char.

The value at which ptr it points is a constant; it will be an error to modify the pointed character; There will not be any error to modify the pointer itself.

Step 4: *ptr = 'a'; Here, we are changing the value of ptr, this will result in the error "cannot modify a const object".

Step 1: const char *s = ""; The constant variable s is declared as an pointer to an array of characters type and initialized with an empty string.

Step 2: char str[] = "Hello"; The variable str is declared as an array of charactrers type and initialized with a string "Hello".

Step 3: s = str; The value of the variable str is assigned to the variable s. Therefore str contains the text "Hello".

Step 4: while(*s){ printf("%c", *s++); } Here the while loop got executed untill the value of the variable s is available and it prints the each character of the variable s.

Hence the output of the program is "Hello".

Step 1: int get(); This is the function prototype for the funtion get(), it tells the compiler returns an integer value and accept no parameters.

Step 2: const int x = get(); The constant variable x is declared as an integer data type and initialized with the value "20".

The function get() returns the value "20".

Step 3: printf("%d", x); It prints the value of the variable x.

Hence the output of the program is "20".

Step 1: const c = -11; The constant variable 'c' is declared and initialized to value "-11".

Step 2: const int d = 34; The constant variable 'd' is declared as an integer and initialized to value '34'.

Step 3: printf("%d, %d\n", c, d); The value of the variable 'c' and 'd' are printed.

Hence the output of the program is -11, 34