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c++compiler-warnings
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Reputation: 1150

Warning C4101 unreferenced local variable

void func(char * param)
{
#ifdef _XYZ_
.....somecode here using param
#endif
.....lots of code here not using param
}

Getting error for compilation and I don't want warning even in code.

Upvotes: 2

Views: 5008

Answers (4)

Alex
Alex

Reputation: 2965

How about: static_cast<void>(param);

Use same method also used to discard result marked [[nodiscard]]:

other than a cast to void, the compiler is encouraged to issue a warning.

[[nodiscard]] bool is_good_idea();

void func(char * param)
{
#ifdef _XYZ_
  .....somecode here using param
#else
  // suppress warning:
  static_cast<void>(param);
#endif

  // also valid:
  static_cast<void>(is_good_idea());
}

Compiler won't produce any additional instructions :)

Upvotes: -1

Remy Lebeau
Remy Lebeau

Reputation: 595782

Prior to C++17's [[maybe_unused]] attribute, you can use compiler-specific commands.

In MSVC, you can do this:

#pragma warning(push)
#pragma warning(disable:4101)
void func(char * param)
{
#ifdef _XYZ_
.....somecode here using param
#endif
}
#pragma warning(pop)

In Borland/Embarcadero, you can do this:

#pragma argsused
void func(char * param)
{
#ifdef _XYZ_
.....somecode here using param
#endif
}

In GCC, you can use one of these:

void func(__attribute__((unused)) char * param)
{
#ifdef _XYZ_
.....somecode here using param
#endif
}

void func(__unused char * param)
{
#ifdef _XYZ_
.....somecode here using param
#endif
}

void func([[gnu::unused]] char * param)
{
#ifdef _XYZ_
.....somecode here using param
#endif
}

Or, you could simply reference the parameter without using any compiler-specific tricks:

void func(char * param)
{
#ifdef _XYZ_
.....somecode here using param
#else
(void)param;
#endif
}

Upvotes: 1

Evg
Evg

Reputation: 26282

Since C++17 we have the maybe_unused attribute to suppress warnings on unused entities:

void func([[maybe_unused]] char * param)
{
#ifdef _XYZ_
.....somecode here using param
#endif
}

Upvotes: 6

Davide Spataro
Davide Spataro

Reputation: 7482

It's simple. If the _XYZ_ is NOT defined the the variable param is not used in the function and the compiler issue a waning that can be translated to:

Translated from compilerian:

My friend you asked me for a piece of memory named param but you are not using it. Are you sure this is intended?

Upvotes: 2

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