Forwarding of return values. Is std::forward is needed?

In the case that you do know that t will not be in a moved-from state after the call to f, your two somewhat sensible options are:

• return std::forward<T>(t) with type T&&, which avoids any construction but allows for writing e.g. auto&& ref = wrapper(42);, which leaves ref a dangling reference

• return std::forward<T>(t) with type T, which at worst requests a move construction when the parameter is an rvalue — this avoids the above problem for prvalues but potentially steals from xvalues

In all cases you need std::forward. Copy elision is not considered because t is always a reference.