c++c++11scopeguardfolly

How to avoid warning when using scope guard?

I am using folly scope guard, it is working, but it generates a warning saying that the variable is unused:

warning: unused variable ‘g’ [-Wunused-variable]

The code:

folly::ScopeGuard g = folly::makeGuard([&] {close(sock);});

How to avoid such warning?

Solution

  • You can disable this warnings by -Wno-unused-variable, though this is a bit dangerous (you loose all realy unused variables).

    One possible solution is to actually use the variable, but do nothing with it. For example, case it to void: 

    (void) g;

    which can be made into a macro:

    #define IGNORE_UNUSED(x) (void) x;

    Alternatively, you can use the boost aproach: declare a templated function that does nothing and use it

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

...

folly::ScopeGuard g = folly::makeGuard([&] {close(sock);});
ignore_unused(g);