Question 1

What will be the output of the program? #include<stdio.h>
int fun(int **ptr);

int main()
{
    int i=10;
    const int *ptr = &i;
    fun(&ptr);
    return 0;
}
int fun(int **ptr)
{
    int j = 223;
    int *temp = &j;
    printf("Before changing p…

Accepted Answer

Error: cannot convert parameter 1 from 'const int **' to 'int **'

Question 2

What will be the output of the program (in Turbo C)? #include<stdio.h>

int fun(int *f)
{
    *f = 10;
    return 0;
}
int main()
{
    const int arr[5] = {1, 2, 3, 4, 5};
    printf("Before modification arr[3] = %d", arr[3]);
    fun(&arr[3]);
    …

Accepted Answer

Step 1: const int arr[5] = {1, 2, 3, 4, 5}; The constant variable arr is declared as an integer array and initialized to arr[0] = 1, arr[1] = 2, arr[2] = 3, arr[3] = 4, arr[4] = 5 Step 2: printf("Before modification arr[3] = %d", arr[3]); It prints the value of arr[3] (ie. 4). Step 3: fun(&arr[3]); The memory location of the arr[3] is passed to fun() and arr[3] value is modified to 10. A const variable can be indirectly modified by a pointer. Step 4: printf("After modification arr[3] = %d", arr[3]); It prints the value of arr[3] (ie. 10). Hence the output of the program is Before modification arr[3] = 4 After modification arr[3] = 10

Question 3

What will be the output of the program? #include<stdio.h>
int get();

int main()
{
    const int x = get();
    printf("%d", x);
    return 0;
}
int get()
{
    return 20;
}

Accepted Answer

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".

Question 4

What will be the output of the program? #include<stdio.h>

int main()
{
    int y=128;
    const int x=y;
    printf("%d
", x);
    return 0;
}

Accepted Answer

Step 1: int y=128; The variable 'y' is declared as an integer type and initialized to value "128". Step 2: const int x=y; The constant variable 'x' is declared as an integer and it is initialized with the variable 'y' value. Step 3: printf("%d
", x); It prints the value of variable 'x'. Hence the output of the program is "128"

Question 5

What will be the output of the program? #include<stdio.h>

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

Accepted Answer

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
", 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.

Question 6

What will be the output of the program in TurboC? #include<stdio.h>
int fun(int **ptr);

int main()
{
    int i=10, j=20;
    const int *ptr = &i;
    printf(" i = %5X", ptr);
    printf(" ptr = %d", *ptr);
    ptr = &j;
    printf(" j = %5X", ptr);…

Accepted Answer

i= FFE4 ptr=10 j=FFE2 ptr=20

Question 7

What will be the output of the program? #include<stdio.h>

int main()
{
    const char *s = "";
    char str[] = "Hello";
    s = str;
    while(*s)
        printf("%c", *s++);

return 0;
}

Accepted Answer

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".

Question 8

What will be the output of the program? #include #include union employee { char name[15]; int age; float salary; }; const union employee e1; int main() { strcpy(e1.name, "K"); printf("%s %d %f", e1.name, e1.a…

Accepted Answer

The output will be (in 16-bit platform DOS): K 75 0.000000

Question 9

What will be the output of the program? #include<stdio.h>

int main()
{
    const int x=5;
    const int *ptrx;
    ptrx = &x;
    *ptrx = 10;
    printf("%d
", x);
    return 0;
}

Accepted Answer

Step 1: const int x=5; The constant variable x is declared as an integer data type and initialized with value '5'. Step 2: const int *ptrx; The constant variable ptrx is declared as an integer pointer. Step 3: ptrx = &x; The address of the constant variable x is assigned to integer pointer variable ptrx. Step 4: *ptrx = 10; Here we are indirectly trying to change the value of the constant vaiable x. This will result in an error. To change the value of const variable x we have to use *(int *)&x = 10;

Question 10

What will be the output of the program? #include<stdio.h>

int main()
{
    const c = -11;
    const int d = 34;
    printf("%d, %d
", c, d);
    return 0;
}

Accepted Answer

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
", c, d); The value of the variable 'c' and 'd' are printed. Hence the output of the program is -11, 34

Question 11

Point out the error in the program. #include #include union employee { char name[15]; int age; float salary; }; const union employee e1; int main() { strcpy(e1.name, "K"); printf("%s", e1.name); e1.ag…

Accepted Answer

Error: cannot modify const object

Question 12

Point out the error in the program. #include<stdio.h>
const char *fun();

int main()
{
    char *ptr = fun();
    return 0;
}
const char *fun()
{
    return "Hello";
}

Accepted Answer

No error and No output

Question 13

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
", array[0], string[0]);
    return…

Accepted Answer

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 14

Point out the error in the program. #include<stdio.h>
const char *fun();

int main()
{
    *fun() = 'A';
    return 0;
}
const char *fun()
{
    return "Hello";
}

Accepted Answer

Error: fun() returns a pointer const character which cannot be modified

Question 15

Point out the error in the program. #include<stdio.h>
#define MAX 128

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

Accepted Answer

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".

Question 16

Point out the error in the program. #include<stdio.h>

int main()
{
    const int x;
    x=128;
    printf("%d
", x);
    return 0;
}

Accepted Answer

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

Question 17

Point out the error in the program. #include<stdio.h>

int main()
{
    const int k=7;
    int *const q=&k;
    printf("%d", *q);
    return 0;
}

Accepted Answer

No error. This will produce 7 as output.

Question 18

Point out the error in the program. #include<stdio.h>
#define MAX 128

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

Accepted Answer

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".

Constants Questions