How do I implement no-op macro (or template) in C++?

Question

How do I implement no-op macro in C++?

#include    

#ifdef NOOP       
    #define conditional_noop(x) what goes here?   
#else       
    #         


        

Answer 1

#ifdef NOOP
    static inline void conditional_noop(int x) { }
#else 
    static inline void conditional_noop(int x) { std::cout << x; }
#endif

Using inline function void enables type checking, even when NOOP isn't defined. So when NOOP isn't defined, you still won't be able to pass a struct to that function, or an undefined variable. This will eventually prevent you from getting compiler errors when you turn the NOOP flag on.

Answer 2

You can just leave it blank. You don't need to follow the #define with anything.

Answer 3

While leaving it blank is the obvious option, I'd go with

#define conditional_noop(x) do {} while(0)

This trick is obviously no-op, but forces you to write a semicolon after conditional_noop(123).

Answer 4

Defining the macro to be void conveys your intent well.

#ifdef NOOP
    #define conditional_noop(x) (void)0
#else

Answer 5

Like others have said, leave it blank.

A trick you should use is to add (void)0 to the macro, forcing users to add a semicolon after it:

#ifdef NOOP       
    #define conditional_noop(x) (void)0
#else       
    #define conditional_noop(x) std::cout << (x); (void)0
#endif  

In C++, (void)0 does nothing. This article explains other not-as-good options, as well as the rationale behind them.

Answer 6

I think that a combination of the previous variants is a good solution:

#ifdef NOOP
    static inline void conditional_noop(int x) do {} while(0)
#else 
    static inline void conditional_noop(int x) do { std::cout << x; } while(0)
#endif

The good thing is that these two codes differ only inside a block, which means that their behaviour for the outside is completely identical for the parser.