Find Output Of Program

Question 1 :

What will be the output of the program?

#include < stdio . h >
int main()
{
    int y=128;
    const int x=y;
    printf("%d\n", x);
    return 0;
}


A). 128
B). Garbage value
C). Error
D). 0
Answer : Option A

Explanation :

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\n", x); It prints the value of variable 'x'.

Hence the output of the program is "128"


Question 2 :

What will be the output of 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 %d %f", e1.name, e1.age, e1.salary);
    return 0;
}


A). Error: RValue required
B). Error: cannot convert from 'const int *' to 'int *const'
C). Error: LValue required in strcpy
D). No error
Answer : Option D

Explanation :

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

K 75 0.000000


Question 3 :

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 ptr = %5x\n", *ptr);
    const *ptr = temp;
    printf("After changing ptr = %5x\n", *ptr);
    return 0;
}


A).
Address of i
Address of j
B).
10
223
C). Error: cannot convert parameter 1 from 'const int **' to 'int **'
D). Garbage value
Answer : Option C

Question 4 :

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\n", x);
    return 0;
}


A). 5
B). 10
C). Error
D). Garbage value
Answer : Option C

Explanation :

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 5 :

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);
    printf(" ptr = %d", *ptr);
    return 0;
}


A). i= FFE2 ptr=12 j=FFE4 ptr=24
B). i= FFE4 ptr=10 j=FFE2 ptr=20
C). i= FFE0 ptr=20 j=FFE1 ptr=30
D). Garbage value
Answer : Option B