What does this if statement check do?

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.