C Programming - Const - Discussion
Discussion Forum : Const - Find Output of Program (Q.No. 5)
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;
}
Discussion:
28 comments Page 2 of 3.
Surbhi said:
1 decade ago
% 5x here x is for hexadecimal value of ptr that has assigned to I and 5 specifies the width of ptr at least 5 digits wide (space also included).
Karun das said:
1 decade ago
Yes, here an object is a constant, we can't change pointer here,
Totally confusion, please correct me if I am thinking in incorrect way.
Totally confusion, please correct me if I am thinking in incorrect way.
Ajay said:
1 decade ago
const int *ptr = &i;
It declares a pointer which points to integer of constant type.
Pointer is not constant.So we can change it.
It declares a pointer which points to integer of constant type.
Pointer is not constant.So we can change it.
Ritesh lenka said:
1 decade ago
Here the int value that pointed by ptr is const. Not the ptr. When we declare int const *ptr, then ptr is consider as a constant.
Aditya said:
7 years ago
What will be the output of the following statement with an explanation?
printf("%X%x%ci%x",11,10,'s',12);
printf("%X%x%ci%x",11,10,'s',12);
Sharath said:
1 decade ago
int i=10;
const int *ptr=10; //correct
*ptr=&i; //correct
*ptr=10; //wrong
const int *ptr=10; //correct
*ptr=&i; //correct
*ptr=10; //wrong
Selvakumar said:
8 years ago
I m not getting the given code. Please anyone explain this.
Himanshu said:
1 decade ago
Can't get it. Please explain it in step by step procedure.
Bhavani said:
9 years ago
How FFE4 and FFE2 came in output? Please explain me.
Pooja said:
6 years ago
I can't understand the output. Please explain me.
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