无操作函数模板

Question

I need to implement a no-op in C++ which preserves all type extensions (const, rvalue ref, lvalue ref). Can I do this as follows?

template<class T>
T && noop (T && t) { return std::forward<T> (t); }

and call it like

... noop (value) ...

(ie use type inference)? Ie does noop (value) have exactly the same type as value ?

Answer 1

Yes it does. However, it's no different from just calling std::forward in the client code itself (except that with std::forward there's no automatic template argument deduction).