Find Output Of Program

Question 6 :

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;
}


A). Error
B). H
C). Hello
D). Hel
Answer : Option C

Explanation :

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 7 :

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;
}


A). Garbage value
B). Error
C). 20
D). 0
Answer : Option C

Explanation :

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 8 :

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]);
    printf("\nAfter modification arr[3] = %d", arr[3]);
    return 0;
}


A).
Before modification arr[3] = 4
After modification arr[3] = 10
B).
Error: cannot convert parameter 1 from const int * to int *
C).
Error: Invalid parameter
D).
Before modification arr[3] = 4
After modification arr[3] = 4
Answer : Option A

Explanation :

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 9 :

What will be the output of the program?

#include < stdio . h >

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


A). 10
B). 11
C). No output
D). Error: ++needs a value
Answer : Option D

Explanation :

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.


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\n", c, d);
    return 0;
}


A). Error
B). -11, 34
C). 11, 34
D). None of these
Answer : Option B

Explanation :

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