Thread: why same address? Need help...

Code:

#include <stdio.h>
int main()
{
  int i;
  char a[]="abc";
  printf("%p %p\n",a[1],i); //different address
  i=a[1];

  printf("%c\n",a[1]);
  printf("%c\n",i);
  printf("%d\n",i);
  printf("%p %p\n",a[1],i); //same address
}

does anybody can explain this? Thx...

Originally Posted by karthigayan

Actually you have to print the address , is it .If you want the print the address of the variables you have to use '&' since you are trying to get the variables address.If you give the pointer there there there is no need for '&'.

Code:

#include <stdio.h>
main()
{
  int i;
  char a[]="abc";
  printf("%p %p\n",&a[1],&i);  // use &
  i=a[1];

  printf("%c\n",a[1]);
  printf("%c\n",i);
  printf("%d\n",i);
  printf("%p %p\n",&a[1],&i);  //use &
}

Result is,

Code:

0xbfd70d9d 0xbfd70da0
b
b
98
0xbfd70d9d 0xbfd70da0

The following is the correct way with out '&'.

Code:

int *p;
printf("%p",p); // No need for & since it is a pointer

I think I know what you mean.
But at first I did't add & in front of the variable, it also worked. what's that address?

I totally understand.
if i don't add & in front of the variable, it prints the value of this variable.
and 62 is binary, change it into decimal. it's 98, it's 'b'.
thank you so much.