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Asked: 2026-06-15T19:59:40+00:00

I’m new to, and learning C++ (know a lot of Java) and the following

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I’m new to, and learning C++ (know a lot of Java) and the following code confuses me…

enter image description here

I know this code fragment is dealing with a pointer-to-function (it’s a callback, that makes sense) but what is throwing me off is the argument between the return type and the function name. What the bloody hell is that?

It looks like a type of function, but I have never heard of that and even after searching and reading about pointer-to-functions I was not able to find anything mentioning that functions could have a type.

If this is true, how does one define a function type?

Thanks, -Cody

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1 Answer

  1. Editorial Team

    GLFWCALL is not a type, it’s a macro which is expanded to a calling convention specific to the platform, or an empty string. Here’s a trimmed fragment of glfw.h:

    #if defined(_WIN32) && defined(GLFW_BUILD_DLL)
 #define GLFWCALL     __stdcall
#elif defined(_WIN32) && defined(GLFW_DLL)
 #define GLFWCALL     __stdcall
#else
 /* We are either building/calling a static lib or we are non-win32 */
 #define GLFWCALL
#endif

    Using a correct calling convention is important on x86/win32, since some of them expect the stack to be cleaned by callee and others by the caller. There can also be differences in the order of passing the arguments.

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