Re: What does CALLBACK mean?

Warren Tang wrote:


You don't. The important thing is that both caller and callee agree on a
calling convention.

  #include <string>
  #include <iostream>
  #include <ostream>

  // declare function type with cdecl
  typedef std::string __cdecl function(std::string str);

  // define function with stdcall
  std::string __stdcall times2( std::string str) {
    return str+str;
  }

  int main() {
    // Note: the cast is necessary to compile this, otherwise
    // the compiler rightfully complains.
    function* pfn = (function*)&times2;

    std::cout << "res=" << (*pfn)("Fou!") << std::endl;
  }

If you run the above program in a debugger, you should get a debug assertion
because the function was called with the incorrect calling convention.
Maybe you have to switch the two calling conventions around, I haven't
actually tested this.

As far as the win32 API is concerned, most function declarations in the
header files include the calling conventions so that changing the default
calling convention doesn't change their meaning.

Uli

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