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

Question

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

#include <iostream>   

#ifdef NOOP       
    #define conditional_noop(x) what goes here?   
#else       
    #define conditional_noop(x) std::cout << (x)   
#endif   
int main() {       
    conditional_noop(123);   
}

I want this to do nothing when NOOP is defined and print "123", when NOOP is not defined.

Answer 1

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

Answer 2

 #ifdef NOOP       
     #define conditional_noop(x)   
 #elif  

nothing!

Answer 3

Defining the macro to be void conveys your intent well.

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

Answer 4

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 5

#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 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.