C ++, preprocessor operator

Question

Can someone explain what this operator does?

#define CONST_SIG (void (*) () ) 1 

Answer 1

This statement defines CONST_SIG as 1 , converted to a pointer to a function that takes no parameters and returns void . This can be useful if you have a pointer to a function, and you might be checking for truth, then CONST_SIG will be true.

You can try the cdecl program, which is available on many linux distributions, to "translate into English" C declarations. Example output in this case:

 cdecl> explain (void (*) () ) cast unknown_name into pointer to function returning void 

Side note: the reason it says "unknown_name" is because our pointer does not have a name. To name it, for example, "p", it will look like this: (void (*p) () ) .

Answer 2

You can pass CONST_SIG to a function that expects a function pointer, and treats the value 1 as a special value for that function pointer.

Answer 3

This link may be relevant http://www.velocityreviews.com/forums/t715007-re-how-to-understand-this-cpp-syntax.html