How to avoid warning when using scope guard?

问题

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?

回答1:

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);

回答2:

You can just label the variable as being unused:

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

Or cast it to void:

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

Neither is great, imo, but at least this lets you keep the warnings.

来源：https://stackoverflow.com/questions/35587076/how-to-avoid-warning-when-using-scope-guard