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Overloading the insertion operator

Hello, I have found something strange. I am using Visual C++ 6.0 SP6 so I am not sure if it is a bug with this or what but here goes.

I wrote a small piece of code:
#include <iostream>

int main()
{
      const char *pMessage = "Hello World\n";
      std::cout.operator<<(pMessage); // Prints out the address of pMessage
      std::cout << pMessage; // Prints out the message

      return 0;
}

When I trace through the code the first call using cout matches the signature as const void * and the second one matches to basic_ostream<_E, _Tr>& _O, const _E *_X in this case: basic_ostream<char, char_traits<char>>& _O, const char *_X as far as I am aware.

What I don't understand is why operator<<() is using a different signature from plain << I thought they are synonymous. Can anyone tell me why? Is it a bug to do with Microsofts C++ compiler?

Thanks
Grant

beryl666

yes. it return the refference of the pMessag only.
see:
http://www.edm2.com/0512/introcpp5.html

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beryl666

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beryl666

std::operator<<(cout,pMessage); //your style

beryl666

std::cout.operator<<(pMessage); // this is only for character
std::cout.operator<<(*(pMessage)); // this u will get character H

//please remember that the style to overload operator are not the same
for example
you want to overload
a++;
++B;
//prefix and postfix it has different style
// if i am not mistaken one is ++ operator() and another is operator()++

hope this clear

beryl666

sorry is operator++()

novitiate

>>// if i am not mistaken one is ++ operator() and another is operator()++
A correction:

operator ++ () and operator ++ (int)

_novi_

beryl666

ok thanks.

SOLUTION

_corey_

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