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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 {
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template <typename Func>
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template <typename Func>
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ScopeExitHelper<Func> ScopeExit(Func&& func) {
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ScopeExitHelper<Func> ScopeExit(Func&& func) {
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return ScopeExitHelper<Func>(std::move(func));
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return ScopeExitHelper<Func>(std::forward<Func>(func));
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}
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}
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} // namespace detail
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} // namespace detail
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