How to define a macro that expands to a conditional statement in C++?

Question

I want to create some context-sensitive macros. The macro

 #define LOG_WARNING(x) qWarning()().noquote().nospace() << x

works fine, it is located in file Macros.h. I want to define a macro, which does not print log messages when called from unit testing routine. So, I modified like

#define LOG_INFO(x) if(!UNIT_TESTING) qInfo().noquote().nospace() << x

Since in this way the macro will depend on UNIT_TESTING, I provided in the same Macros.h

extern bool UNIT_TESTING;   // Whether in course of unit testing

However, the compilers tells

declaration does not declare anything [-fpermissive]
 extern bool UNIT_TESTING; // Whether in course of unit testing
        ^

At the same time, if the external is declared in the file from which Macros.h is included, it works fine. Do I wrong something?

Are you in a `extern "C"` block ? Is `UNIT_TESTING` a macro ? — Jarod42, Aug 25 '17 at 12:03
This of course is not a conditional macro. It's an unconditional macro that expands to a conditional statement. — MSalters, Aug 25 '17 at 12:08
Are you really sure you want to use a macro? A logging class is a lot more bullet proof and you can feed that class an attribute on when to print and when not to print. — NathanOliver, Aug 25 '17 at 12:10
@Jarod42 UNIT_TESTING is a simple bool, and not in extern "C" — katang, Aug 25 '17 at 12:24

score 2 · Accepted Answer · answered Aug 25 '17 at 12:12

Here is how to share variables across source files. Nevertheless, I would highly recommend not to do so, but to implement a function (bool IS_UNIT_TESTING() ) or class which takes care of this. In this way you can change the implementation without changing the interface.

Moreover, Macros are evil. They are error prone can not be debuged easily. Use inline functions or constexp instead. The compiler will optimize it to almost the very same code.

How to define a macro that expands to a conditional statement in C++?

1 Answers1