common/scope_exit: Replace std::move with std::forward in ScopeExit()
The template type here is actually a forwarding reference, not an rvalue reference in this case, so it's more appropriate to use std::forward to preserve the value category of the type being moved.
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		| @@ -20,7 +20,7 @@ struct ScopeExitHelper { | ||||
|  | ||||
| template <typename Func> | ||||
| ScopeExitHelper<Func> ScopeExit(Func&& func) { | ||||
|     return ScopeExitHelper<Func>(std::move(func)); | ||||
|     return ScopeExitHelper<Func>(std::forward<Func>(func)); | ||||
| } | ||||
| } // namespace detail | ||||
|  | ||||
|   | ||||
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