如果语句检查该怎么办？

Question

In the following code the programmer puts the name of a function inside an if condition as if to check something before going ahead and registering a callback:

#if _DEBUG
    if(glDebugMessageCallback) // He checks this function, presumably it returns true if it exists
{
        cout << "Register OpenGL debug callback " << endl;
        glEnable(GL_DEBUG_OUTPUT_SYNCHRONOUS);
        glDebugMessageCallback(openglCallbackFunction, nullptr); // Calls the function here
        GLuint unusedIds = 0;
        glDebugMessageControl(GL_DONT_CARE,
            GL_DONT_CARE,
            GL_DONT_CARE,
            0,
            &unusedIds,
            true);
    }
    else
        cout << "glDebugMessageCallback not available" << endl;  // So if the if condition evaluated to false, the function doesn't exist.
#endif

My question is why this way of checking if a function exists? If a function doesn't exist surely you'll get a compile error telling you that the function doesn't exist, this seems strange to me. I know basically the function address evaluates to bool, and I asked a question before about this and was told that function addresses aren't particularly useful as implicit conversion to bool, and I don't see how.

I should mention also that the function inside the if condition is a MACRO define, defined as:

#define glDebugMessageCallback GLEW_GET_FUN(__glewDebugMessageCallback)

If it being a define macro makes a difference.

Answer 1

glDebugMessageCallback is either going to be null (0), or the (presumably valid) address of a function that matches the signature required.

The if statement will evaluate to true for any non-null value, and false for a null value. (there may be a compiler warning, but the code will work.)

So, if glDebugMessageCallback was initialized to null, and never set to a function address, the if statement will evaluate to false. If glDebugMessageCallback was set to the address of a function, the code inside the if block will execute, and the function will be called.