regarding const pointer

Code:

int i=10;
	const int ci=123;
	const int *cpi;
	int *ncpi;
	cpi=&ci;
	printf("Main cpi  %d\n",*cpi);
	ncpi=&i;
	printf("Main ncpi %d\n",*ncpi);
	cpi=ncpi;
	printf("second %d\n",*cpi); 
	ncpi=(int *)cpi;
	printf("%d\n",*ncpi);
	[B]*ncpi=0;[/B]
	exit();

My question is why *ncpi =0 ,

 

Can you post a complete program, including
- the main(),
- the include files,
- how you compiled it,
- and what you saw on the output (copy/paste from your console).

For instance, C and C++ handle 'const' differently.

 

Code:

#include<stdio.h>
main()
{
	int i=10;
	const int ci=123;
	const int *cpi;
	int *ncpi;
	cpi=&ci;
	printf("Main cpi  %d\n",*cpi);
	ncpi=&i;
	printf("Main ncpi %d\n",*ncpi);
	cpi=ncpi;
	printf("second %d\n",*cpi); 
	ncpi=(int *)cpi;
	printf("%d\n",*ncpi);
	*ncpi=0;
	exit();
}

output

Code:

Main cpi  123
Main ncpi 10
second 10 
10

if it dont include *ncpi=0 it gives the same result

 

Why would attempting the assignment AFTER the value has been printed have any effect whatsoever?

Does exit(); really compile for you?

 

its a book example ,thats why i asked why it is required to do so ,even i couldnt get that