Forward individual members of a Forward reference

问题

I have a function that potentially moves a generic argument but through their members. What of these options is more correct:

1. This seems the more natural but it is strange because the argument is potentially moved twice [a], which is odd because the object can become invalid.

   template<class T> 
   void fun(T&& t){
       myhead_ = std::forward<T>(t).head_;
       myrest_ = std::forward<T>(t).rest_;
   }
   

2. This can't be incorrect but it may not be moving anything.

   template<class T> void fun(T&& t){
       myhead_ = std::forward<decltype(t.head_)>(t.head_);
       myrest_ = std::forward<decltype(t.rest_)>(t.rest_);
   }
   

3. This seems correct but too much code.

   template<class T> void fun(T& t){
       myhead_ = t.head_;
       myrest_ = t.rest_;
   }
   template<class T> void fun(T&& t){
       myhead_ = std::move(t.head_);
       myrest_ = std::move(t.rest_);
   }
   

---

[a] This statement is incorrect as @Angew pointed out, it only looks as if it is moved twice. std::forward (like std::move) doesn't actually move anything. At most the member is moved (by the subsequent operation decltype(myhead)::operator= but that is precisely the objective.)

回答1:

Your first code is perfectly fine:

template<class T> 
void fun(T&& t){
    myhead_ = std::forward<T>(t).head_;
    myrest_ = std::forward<T>(t).rest_;
}

That is because the standard guarantees that when doing a.b and a is an xvalue (such as a forwarded rvalue reference), the result of a.b is also an exvalue (i.e. can be moved from). Also note that std::forward and std::move do not do any actual moving themselves, they're just casts. So there is no risk in moving from t twice in your code.

来源：https://stackoverflow.com/questions/50149972/forward-individual-members-of-a-forward-reference