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1.
What will be the output of the program?

#include
int main()
{
int y=128;
const int x=y;
printf("%d\n", x);
return 0;
}
Answer & Solution
Answer: (C) 128
Solution:
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"
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2.
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.age, e1.salary);
return 0;
}
Answer & Solution
Answer: (D) No Error
Solution:
The output will be (in 16-bit platform DOS):
K 75 0.000000
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3.
What will be the output of the program?

#include
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;
}
Answer & Solution
Answer: (C) Error: cannot convert parameter 1 from 'const int **' to 'int **'
Solution:
No answer description available for this question. Let us discuss. close
4.
What will be the output of the program?

#include
int main()
{
const int x=5;
const int *ptrx;
ptrx = &x;
*ptrx = 10;
printf("%d\n", x);
return 0;
}
Answer & Solution
Answer: (C) Error
Solution:
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;v
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5.
What will be the output of the program in TurboC?

#include
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;
}
Answer & Solution
Answer: (B) i= FFE4 ptr=10 j=FFE2 ptr=20
Solution:
No answer description available for this question. Let us discuss.
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