address of a character pointer

[swethapradeep]

Hi,

i was trying to access the address of the character pointer it gives me the values stored in the variable.

#include<iostream>
using namespace std;
int  main()
{
  char *i;
   i=new char[2];
   i[0]='1';
   i[1]='2';
   cout<<i<<"\n";
  delete[]i;
 return 0;
}

now my output is:

12

then i changed the type of the variable to int,

#include<iostream>
using namespace std;
int  main()
{
  int *i;
   i=new int[2];
   i[0]=1;
   i[1]=2;
   cout<<i<<"\n";
  delete[]i;
 return 0;
}

the out put is the address which is expected

0x9d99008

why is my code not giving the address of the variable i when it is a character pointer.
-swetha

[GNU-Fan]

why is my code not giving the address of the variable i when it is a character pointer.

It is the other way around,

i

is refering to the address of the variable both times.

In the case of 'char',

type(i) = char*

If the function cout gets such a char*, it automatically[*] knows that it should display the null-terminated string beginning at this address. (You forgot the terminating 0 by the way, making it crash-prone.)

[*] It knows because the operator << is overloaded for several different types on the right hand side.

[gerard4143]

If you want the address of the elements of a character array try:

int x = 0;
char *i = new char[2];


cout<<&i[x]<<"\n";