What is the best way to suppress “Unused variable x”-warning [duplicate]

Answer 1

(void) variable might work for some compilers.

For C++ code, also see http://herbsutter.com/2009/10/18/mailbag-shutting-up-compiler-warnings/ where Herb Sutter recommends using:

template<class T> void ignore( const T& ) { }

...

ignore(variable);

Answer 2

Delete the unused variable declaration from the code. (pun intended)

(What??? it's what I do: point that the obvious is most likely the best solution.)

Now, from comments on other answers, apparently it's garbage generated from macros. Well, that's a pleonasm.

Solutions:

• refactor that macro to #if declare the variable only if it's really used;
• create another version of the macro that skips the unused variable generation.
• Better still, avoid using macros that bring issues to the code.

Answer 3

You can silence the warning using #pragma

#pragma clang diagnostic push
#pragma clang diagnostic ignored "-Wunused"

int unususedVariable = 1;

#pragma clang diagnostic pop

If you are using GCC, use #pragma gcc ...

Answer 4

If its used and you are shipping the project, delete it. Worst comment it.

Answer 5

Found an article http://sourcefrog.net/weblog/software/languages/C/unused.html that explains UNUSED. Interesting that the author also mangles the unused variable name so you can't inadvertently use it in the future.

Excerpt:

#ifdef UNUSED
#elif defined(__GNUC__)
# define UNUSED(x) UNUSED_ ## x __attribute__((unused))
#elif defined(__LCLINT__)
# define UNUSED(x) /*@unused@*/ x
#else
# define UNUSED(x) x
#endif

void dcc_mon_siginfo_handler(int UNUSED(whatsig))