__w64
The latest version of this topic can be found at __w64.
Microsoft Specific) This keyword is obsolete. In versions of Visual Studio earlier than Visual Studio 2013, this lets you mark variables, so that when you compile with /Wp64 the compiler will report any warnings that would be reported if you were compiling with a 64-bit compiler.
Syntax
type __w64 identifier
Parameters
type
One of the three types that could cause problems in code being ported from a 32-bit to a 64-bit compiler: int, long, or a pointer.
identifier
The identifier for the variable you are creating.
Remarks
Important
The /Wp64 compiler option and __w64 keyword are deprecated in Visual Studio 2010 and Visual Studio 2013 and removed starting in Visual Studio 2013. If you use the /Wp64 compiler option on the command line, the compiler issues Command-Line Warning D9002. The __w64 keyword is silently ignored. Instead of using this option and keyword to detect 64-bit portability issues, use a Visual C++ compiler that targets a 64-bit platform. For more information, see Configuring Programs for 64-Bit.
Any typedef that has __w64 on it must be 32 bits on x86 and 64 bits on x64.
To detect portability issues by using versions of the Visual C++ compiler earlier than Visual Studio 2010, the __w64 keyword should be specified on any typedefs that change size between 32 bit and 64 bit platforms. For any such type, __w64 must appear only on the 32-bit definition of the typedef.
The __w64 keyword is ignored if the compilation does not use /Wp64.
For more information about porting to 64-bit, see the following topics:
Example
// __w64.cpp
// compile with: /W3 /Wp64
typedef int Int_32;
#ifdef _WIN64
typedef __int64 Int_Native;
#else
typedef int __w64 Int_Native;
#endif
int main() {
Int_32 i0 = 5;
Int_Native i1 = 10;
i0 = i1; // C4244 64-bit int assigned to 32-bit int
// char __w64 c; error, cannot use __w64 on char
}