Does Q_UNUSED have any side effects?

Question

Given the following piece of code:

void test(int var)
{
     Q_UNUSED(var);
#ifdef SOMETHING
     printf("%d",var);
     //do something else with var...
#endif
}

Would the Q_UNUSED macro have any effect if I actually use the 'var' variable in some scenario (like in the example above), or it has no effect at all when I suppress compiler warnings for unused variables?

So far I observe it has no effect, but I would like to make sure.

Solution

No in many cases (e.g. just passing a simple variable to the macro). The definition is inside qglobal.h:

#  define Q_UNUSED(x) (void)x;

To disable unused variable warnings. You can use the variable after this macro without any problem.

However, if you pass an expression or something else to the macro and the compiler has to evaluate the expression it may has side effects^† .

Comments

[Admin]

Just look at its documentation. If there's none, then read its definition. (hint: it probably hasn't any. It most certainly cannot possibly render a variable "unusable". I guess it's something like ((void)(expression));

[MSalters]

@H2CO3: Couldn't it redeclare var to make any subsequent use ambiguous? extern qUnusedType var;

[László Papp]

@H2CO3: correct, it is not that magical.

[Mikhail]

Possible duplicate of Why cast unused return values to void?

[HostileFork says dont trust SE]

I posted an "UNUSED with teeth" on CodeReview SE... for anyone who is interested in the idea of actually corrupting the data. (Not ideal to corrupt it but the toothlessness of unused annotation was continued even in standard C++ annotations, which puzzlingly went with [[maybe_unused]] on the parameter definition, instead of letting you mark points in the control flow after which you don't want a variable used and have the compiler catch it.)

[László Papp]

The only side effect, I can see, is the porting effort if you switch away from Qt, but that is not a big deal since you would have more issues anyway.