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Old 01-28-2010, 05:38 AM   #1
ishandutta2007
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Registered: May 2009
Location: kolkata,india
Distribution: fedora
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Explain the output(pointers)


explain me the output please .As far as my knowledge. a is a character type pointer variable and its size should be 4 bytes irrespective of what its pointing to.but i get a result =50.
*a is the value of a[0] and its size is rightly 1 byte.
but how can the size of that variable change to 4 byte when i just add 7 to it(although i don't store the added result anywhere)



Code:
#include<stdio.h>
int main(){
	char a[50];
	printf("\n%d  ",sizeof(a));
	printf("\n%d  ",sizeof(*a));
	printf("\n%d  ",sizeof(*a+7));
	return 0;
	}
output:
Code:
50
1
4

Last edited by ishandutta2007; 01-28-2010 at 05:42 AM.
 
Old 01-28-2010, 06:19 AM   #2
SigTerm
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Registered: Dec 2009
Distribution: Slackware 12.2
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Quote:
Originally Posted by ishandutta2007 View Post
but i get a result =50.
"a" is an array of 50 chars, so size is correct.
Try this:
Code:
#include<stdio.h>
int main(){
	char a[50];
	char* b = a;
	printf("\n%d  ",sizeof(a));
	printf("\n%d  ",sizeof(b));
	return 0;
}

Last edited by SigTerm; 01-28-2010 at 09:09 AM.
 
Old 01-28-2010, 08:25 AM   #3
Dan04
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Distribution: Ubuntu
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*a+7 means (*a)+7. And char+int=int.
 
1 members found this post helpful.
  


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