How to declare 64 bits in Visual Studio 6

Did you ever needed to declare a 64 bits integer using Visual Studio 6?

On newer versions of Visual Studio and on gcc you can use the ll to tell the compiler that the value is 64 bits wide. But Visual Studio 6 does not understand it.

How do you do? You must use the keyword

__int64.

I discovered it because a project I wrote using Visual Studio 2005 that must be compiled using Visual Studio 6.

Below you can see a sample code that compiles both in VS 6 and 2005.

#include 
#include "stdafx.h"
#include 

int main(int argc, char* argv[])
{
  __int64 i = 0;

  printf("Hello World! %08x %08x\n", i);

  #ifdef _MSC_VER
      #if _MSC_VER <= 1200
         // Visual Studio 6 or lower do not accept ll as 64 bits constants.
         #define __IGNORE_LL_MODIFIER_MS         
         #endif // _MSC_VER     
      #endif // _MSC_VER

  #ifdef __IGNORE_LL_MODIFIER_MS
      i = ((__int64)0x1234567890123456);
  #else
      i = 0x1234567890123456ll;
  #endif

  printf("Hello World! %08x %08x\n", i);
  getchar();
  return 0;
}

I really hope this piece of information will be helpful to someone.