Question 1

Point out the error in the program.

#include<stdio.h>
#include<stdlib.h>

union employee
{
    char name[15];
    int age;
    float salary;
};
const union employee e1;

int main()
{
    strcpy(e1.name, "K");
    printf("%s", e1.name);    
    e1.age=85;
    printf("%d", e1.age);
    printf("%f", e1.salary);
    return 0;
}

Answer

Answer: B

No answer description is available. Let's discuss.

Question 2

Point out the error in the program (in Turbo-C).

#include<stdio.h>
#define MAX 128

int main()
{
    const int max=128;
    char array[max];
    char string[MAX];
    array[0] = string[0] = 'A';
    printf("%c %c\n", array[0], string[0]);
    return 0;
}

Answer

Answer: A

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.

Question 3

Point out the error in the program.

#include<stdio.h>
#define MAX 128

int main()
{
    char mybuf[] = "India";
    char yourbuf[] = "BIX";
    char *const ptr = mybuf;
    *ptr = 'a';
    ptr = yourbuf;
    return 0;
}

Answer

Answer: B

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

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

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".

Question 4

What will be the output of the program?

#include<stdio.h>

int main()
{
    const c = -11;
    const int d = 34;
    printf("%d, %d\n", c, d);
    return 0;
}

Answer

Answer: B

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

Question 5

What will be the output of the program?

#include<stdio.h>

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

Answer

Answer: D

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.

Questions for: Const