How do I best silence a warning about unused variables?

Question

I have a cross platform application and in a few of my functions not all the values passed to functions are utilised. Hence I get a warning from GCC telling me that there are unused variables.

What would be the best way of coding around the warning?

An #ifdef around the function?

#ifdef _MSC_VER
void ProcessOps::sendToExternalApp(QString sAppName, QString sImagePath, qreal qrLeft, qreal qrTop, qreal qrWidth, qreal qrHeight)
#else
void ProcessOps::sendToExternalApp(QString sAppName, QString sImagePath, qreal /*qrLeft*/, qreal /*qrTop*/, qreal /*qrWidth*/, qreal /*qrHeight*/)
#endif
{

This is so ugly but seems like the way the compiler would prefer.

Or do I assign zero to the variable at the end of the function? (which I hate because it's altering something in the program flow to silence a compiler warning).

Is there a correct way?

Answer 1

You can put it in "(void)var;" expression (does nothing) so that a compiler sees it is used. This is portable between compilers.

E.g.

void foo(int param1, int param2)
{
    (void)param2;
    bar(param1);
}

Or,

#define UNUSED(expr) do { (void)(expr); } while (0)
...

void foo(int param1, int param2)
{
    UNUSED(param2);
    bar(param1);
}

Answer 2

In GCC and Clang you can use the __attribute__((unused)) preprocessor directive to achieve your goal.
For example:

int foo (__attribute__((unused)) int bar) {
   return 0;
}

Answer 3

C++17 now provides the [[maybe_unused]] attribute.

http://en.cppreference.com/w/cpp/language/attributes

Quite nice and standard.

Answer 4

C++17 Update

In C++17 we gain the attribute [[maybe_unused]] which is covered in [dcl.attr.unused]

The attribute-token maybe_unused indicates that a name or entity is possibly intentionally unused. It shall appear at most once in each attribute-list and no attribute-argument-clause shall be present. ...

Example:

 [[maybe_unused]] void f([[maybe_unused]] bool thing1,
                        [[maybe_unused]] bool thing2) {
  [[maybe_unused]] bool b = thing1 && thing2;
    assert(b);
 }

Implementations should not warn that b is unused, whether or not NDEBUG is defined. —end example ]

For the following example:

int foo ( int bar) {
    bool unused_bool ;
    return 0;
}

Both clang and gcc generate a diagnostic using -Wall -Wextra for both bar and unused_bool (See it live).

While adding [[maybe_unused]] silences the diagnostics:

int foo ([[maybe_unused]] int bar) {
    [[maybe_unused]] bool unused_bool ;
    return 0;
}

see it live.

Before C++17

In C++11 an alternative form of the UNUSED macro could be formed using a lambda expression(via Ben Deane) with an capture of the unused variable:

#define UNUSED(x) [&x]{}()

The immediate invocation of the lambda expression should be optimized away, given the following example:

int foo (int bar) {
    UNUSED(bar) ;
    return 0;
}

we can see in godbolt that the call is optimized away:

foo(int):
xorl    %eax, %eax
ret

Answer 5

Your current solution is best - comment out the parameter name if you don't use it. That applies to all compilers, so you don't have to use the pre-processor to do it specially for GCC.

Answer 6

An even cleaner way is to just comment out variable names:

int main(int /* argc */, char const** /* argv */) {
  return 0;
}

Answer 7

gcc doesn't flag these warnings by default. This warning must have been turned on either explicitly by passing -Wunused-parameter to the compiler or implicitly by passing -Wall -Wextra (or possibly some other combination of flags).

Unused parameter warnings can simply be suppressed by passing -Wno-unused-parameter to the compiler, but note that this disabling flag must come after any possible enabling flags for this warning in the compiler command line, so that it can take effect.

Answer 8

A coworker just pointed me to this nice little macro here

For ease I'll include the macro below.

#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))

Answer 9

macro-less and portable way to declare one or more parameters as unused:

template <typename... Args> inline void unused(Args&&...) {}

int main(int argc, char* argv[])
{
    unused(argc, argv);
    return 0;
}

Answer 10

I have seen this instead of the (void)param2 way of silencing the warning:

void foo(int param1, int param2)
{
    std::ignore = param2;
    bar(param1);
}

Looks like this was added in C++11

Answer 11

Lol! I dont think there is another question on SO that reveal all the heretics corrupted by Chaos better that this one!

With all due respect to C++17 there is a clear guideline in C++ Core Guidelines. AFAIR, back in 2009 this option was available as well as today. And if somebody says it is considered as a bug in Doxygen then there is a bug in Doxygen

Answer 12

Using preprocessor directives is considered evil most of the time. Ideally you want to avoid them like the Pest. Remember that making the compiler understand your code is easy, allowing other programmers to understand your code is much harder. A few dozen cases like this here and there makes it very hard to read for yourself later or for others right now.

One way might be to put your parameters together into some sort of argument class. You could then use only a subset of the variables (equivalent to your assigning 0 really) or having different specializations of that argument class for each platform. This might however not be worth it, you need to analyze whether it would fit.

If you can read impossible templates, you might find advanced tips in the "Exceptional C++" book. If the people who would read your code could get their skillset to encompass the crazy stuff taught in that book, then you would have beautiful code which can also be easily read. The compiler would also be well aware of what you are doing (instead of hiding everything by preprocessing)

Answer 13

First off the warning is generated by the variable definition in the source file not the header file. The header can stay pristine and should, since you might be using something like doxygen to generate the API-documentation.

I will assume that you have completely different implementation in source files. In these cases you can either comment out the offending parameter or just write the parameter.

Example:

func(int a, int b)
{
    b;
    foo(a);
}

This might seem cryptic, so defined a macro like UNUSED. The way MFC did it is:

#ifdef _DEBUG
#define UNUSED(x)
#else
#define UNUSED(x) x
#endif

Like this you see the warning still in debug builds, might be helpful.

Answer 14

Is it not safe to always comment out parameter names? If it's not you can do something like

#ifdef _MSC_VER
# define P_(n) n
#else
# define P_(n)
#endif

void ProcessOps::sendToExternalApp(
    QString sAppName, QString sImagePath,
    qreal P_(qrLeft), qreal P_(qrTop), qreal P_(qrWidth), qreal P_(qrHeight))

It's a bit less ugly.

Answer 15

Using an UNREFERENCED_PARAMETER(p) could work. I know it is defined in WinNT.h for Windows systems and can easily be defined for gcc as well (if it doesn't already have it).

UNREFERENCED PARAMETER(p) is defined as

#define UNREFERENCED_PARAMETER(P)          (P)

in WinNT.h.

Answer 16

Use compiler's flag, e.g. flag for GCC: -Wno-unused-variable

Answer 17

This works well but requires C++11

template <typename ...Args>
void unused(Args&& ...args)
{
  (void)(sizeof...(args));
}

Answer 18

In C++11, this is the solution I'm using:

template<typename... Ts> inline void Unreferenced(Ts&&...) {}

int Foo(int bar) 
{
    Unreferenced(bar);
    return 0;
}

int Foo2(int bar1, int bar2) 
{
    Unreferenced(bar1, bar2);
    return 0;
}

Verified to be portable (at least on modern msvc, clang and gcc) and not producing extra code when optimizations are enabled. With no optimization, the extra function call is performed and references to the parameters are copied to the stack, but there are no macros involved.

If the extra code is an issue, you can use this declaration instead:

(decltype(Unreferenced(bar1, bar2)))0;

but at that point, a macro provides better readability:

#define UNREFERENCED(...) { (decltype(Unreferenced(__VA_ARGS__)))0; }

Answer 19

I found most of the presented answers work for local unused variable only, and will cause compile error for unused static global variable.

Another macro needed to suppress the warning of unused static global variable.

template <typename T>
const T* UNUSED_VARIABLE(const T& dummy) { 
    return &dummy;
}
#define UNUSED_GLOBAL_VARIABLE(x) namespace {\
    const auto dummy = UNUSED_VARIABLE(x);\
}

static int a = 0;
UNUSED_GLOBAL_VARIABLE(a);

int main ()
{
    int b = 3;
    UNUSED_VARIABLE(b);
    return 0;
}

This works because no warning will be reported for non-static global variable in anonymous namespace.

C++ 11 is required though

 g++  -Wall -O3  -std=c++11 test.cpp

Answer 20

void func(void *aux UNUSED)
{
    return;
}

smth like that, in that case if u dont use aux it wont warn u

Answer 21

define a macro for unused local variable:

#define UNUSED(x)  (void)((x))

Answer 22

You can use __unused to tell the compiler that variable might not be used.

- (void)myMethod:(__unused NSObject *)theObject    
{
    // there will be no warning about `theObject`, because you wrote `__unused`

    __unused int theInt = 0;
    // there will be no warning, but you are still able to use `theInt` in the future
}

Answer 23

I don't see your problem with the warning. Document it in the method/function header that compiler xy will issue a (correct) warning here, but that theses variables are needed for platform z.

The warning is correct, no need to turn it off. It does not invalidate the program - but it should be documented, that there is a reason.