Supress unused variable warning in C++ => Compiler bug or code bug?

Question

Presently, I am using the following function template to suppress unused variable warnings:

template<typename T>
void
unused(T const &) {
  /* Do nothing. */
}

However, when porting to cygwin from Linux, I am now getting compiler errors on g++ 3.4.4 (On linux I am 3.4.6, so maybe this is a bug fix?):

Write.cpp: In member function `void* Write::initReadWrite()':
Write.cpp:516: error: invalid initialization of reference of type 'const volatile bool&' from expression of type 'volatile bool'
../../src/common/Assert.h:27: error: in passing argument 1 of `void unused(const T&) [with T = volatile bool]'
make[1]: *** [ARCH.cygwin/release/Write.o] Error 1

The argument to unused is a member variable declared as:

  volatile bool readWriteActivated;

Is this a compiler bug or a bug in my code?

Here is the minimal test case:

template<typename T>
void unused(T const &) { }

int main() {
  volatile bool x = false;
  unused(!x); // type of "!x" is bool
}

Answer 1

In GCC, you can define a macro as follows:

#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

Any parameters marked with this macro will suppress the unused warning GCC emits (and renames the parameter with a prefix of UNUSED_). For Visual Studio, you can suppress warnings with a #pragma directive.

Answer 2

I'm not 100% sure that this is portable, but this is the idiom I've usually used for suppressing warnings about unused variables. The context here is a signal handler that is only used to catch SIGINT and SIGTERM, so if the function is ever called I know it's time for the program to exit.

volatile bool app_killed = false;
int signal_handler(int signum)
{
    (void)signum; // this suppresses the warnings
    app_killed = true;
}

I tend to dislike cluttering up the parameter list with __attribute__((unused)), since the cast-to-void trick works without resorting to macros for Visual C++.

Answer 3

The actual way of indicating you don't actually use a parameter is not giving it a name:

int f(int a, float) {
     return a*2;
}

will compile everywhere with all warnings turned on, without warning about the unused float. Even if the argument does have a name in the prototype (e.g. int f(int a, float f);), it still won't complain.

Answer 4

It is a compiler bug and there are no known work arounds:

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=42655

Answer 5

The answer proposed by haavee (amended by ur) is the one I would normally use:

int f(int a, float /*epsilon*/) {
     return a*2;
}

The real problem happens when the argument is sometimes but not always used in the method, e.g.:

int f(int a, float epsilon) {
#ifdef LOGGING_ENABLED
     LOG("f: a = %d, epsilon = %f\n", a, epsilon);
#endif
     return a*2;
}

Now, I can't comment out the parameter name epsilon because that will break my logging build (I don't want to insert another #ifdef in the argument list because that makes the code much harder to read).

So I think the best solution would be to use Tom's suggestion:

int f(int a, float epsilon) {
(void) epsilon;    // suppress compiler warning for possibly unused arg
#ifdef LOGGING_ENABLED
     LOG("f: a = %d, epsilon = %f\n", a, epsilon);
#endif
     return a*2;
}

My only worry would be that some compilers might warn about the "(void) epsilon;" statement, e.g. "statement has no effect" warning or some such - I guess I'll just have to test on all the compilers I'm likely to use...