CODEWITHSUNDEEP
c++macros
gsf
gsf

Reputation: 7242

How do you disable the unused variable warnings when you are not sure that what you have is variable

There are many different techniques to disable unused variable warnings. Some with void, same with attribute. They all though assume that you know that it is a variable.

I have a case of macro that changes its behavior from release and debug, say:

#if NDEBUG
#define MACRO(X)
#else
#define MACRO(X) do_something(X)
#endif

Code like:

void foo(int a) {
    MACRO(a);
}

may result in warning in release. I would like to change this to:

#if NDEBUG
#define MACRO(X) UNUSED(X)
#else
#define MACRO(X) do_something(X)
#endif

So the question is what I should define UNUSED to, when the task is complicated from the fact that the argument to MACRO is not limited to variable, but it could also be a function call or any other expression. Or

MACRO(5);
MACRO(a+b);
MACRO(foo());

are also valid uses of MACRO

Upvotes: 1

Views: 707

Answers (2)

Basile Starynkevitch
Basile Starynkevitch

Reputation: 1

You might have:

#define UNUSED(X) (false && ((X),true))

and-then with comma-operator (for typing reasons)

Or

#define UNUSED(X) do{if (false) (void)(X);}while(0)

always false conditional

Upvotes: 1

Simple
Simple

Reputation: 14390

Here's an example I like:

#ifdef NDEBUG
#define MACRO(X) ((void)sizeof((X), 0))
#else
#define MACRO(X) do_something((X))
#endif

The names you use in the macro argument end up being used, but in an unevaluated context and then that gets cast to void which prevents a different warning.

Upvotes: 1

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