a bit confused with char string~

[black]

as we know something like this would get the value of the pointer: *pt
and this would get the adress what the pointer point to: &pt

but when I experiment this to char strings I got the different response.

for example:

void main()
{
char arr[]="Hello World";
cout << &arr << endl;
}

Let's check what we get.

[Unregistered]

Originally posted by black
void main()...

Let's check what we get.

We get slapped on the wrist for using void main.

[ygfperson]

arr is a pointer
&arr is the address of that pointer variable
*arr is the value of the char the pointer points to

what are you trying to do with that code?
i got 0xbffffa4c if you're interested

[Prelude]

#include <iostream>

int main()
{
  char arr[] = "Hello World";

  // This should print an address
  std::cout<< &arr <<std::endl;
  // This should print the contents of arr
  // ie. Hello World
  std::cout<< arr <<std::endl;
  // This should print H
  std::cout<< *arr <<std::endl;
  
  return 0;
}

So &arr does what we expect, it returns the address of the first element of the array. *arr does as we expect as well, it returns the value of the first element of the array. Since cout assumes that you want to print the array when you give it arr, it does so instead of printing the contents of arr, which is an address.

>void main()
I've explained this to you before, what's your excuse for still using it?

-Prelude