the expression "*root" causes the structure to be printed,which
prints,(root->data=),10,(root->left=)0,(root->right)=0
and then the last value is &(root->data),other remaining values are ignored

On Fri, Jul 15, 2011 at 12:16 AM, sagar pareek <[email protected]>wrote:

> #include<stdio.h>
> #include<stdlib.h>
> struct node
> {
>     int data;
>     struct node *left;
>     struct node *right;
> }*root=NULL;
> main()
> {
>     root=(struct node *)malloc(sizeof(struct node));
>     root->data=10;
>     root->left=NULL;
>     root->right=NULL;
>
> printf("&root->%x--root-->%x--*root-->%d--&(root->data)=%x--&(root->left)=%x--&(root-->right)=%x",&root,root,*root,&(root->data),&(root->left),&(root->right));
> }
>
> In this,program,i try to print address of struct members but it print zero
> for &(root->data) and &(root->left).it also print same value for root and
> &(root->right).so anyone one can explain me in detail how struct members
> stored in memory.
>
> --
> **Regards
> SAGAR PAREEK
> COMPUTER SCIENCE AND ENGINEERING
> NIT ALLAHABAD
>
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-- 
*Regards,*
*Piyush Kapoor,*
*CSE-IT-BHU*

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