C++ universal reference in constructor and return value optimization (rvo)

问题

Why does rvalue optimization not occur in classes with constructor with universal reference arguments?

http://coliru.stacked-crooked.com/a/672f10c129fe29a0

#include <iostream>

 template<class ...ArgsIn>
struct C {

  template<class ...Args>
  C(Args&& ... args) {std::cout << "Ctr\n";}        // rvo occurs without &&

  ~C(){std::cout << "Dstr\n";}
};

template<class ...Args> 
auto f(Args ... args) {
    int i = 1;
  return C<>(i, i, i);
}

int main() {
  auto obj = f();
}

Output:

Ctr
Ctr
Dstr
Ctr
Dstr
Dstr

回答1:

I believe that the problem is that instantiations of

template<class ...Args>
C(Args&& ... args) {std::cout << "Ctr\n";}  

are not copy/move constructors as far as the language is concerned and therefore calls to them cannot be elided by the compiler. From §12.8 [class.copy]/p2-3, emphasis added and examples omitted:

A non-template constructor for class X is a copy constructor if its first parameter is of type X&, const X&, volatile X& or const volatile X&, and either there are no other parameters or else all other parameters have default arguments (8.3.6).

A non-template constructor for class X is a move constructor if its first parameter is of type X&&, const X&&, volatile X&&, or const volatile X&&, and either there are no other parameters or else all other parameters have default arguments (8.3.6).

In other words, a constructor that is a template can never be a copy or move constructor.

The return value optimization is a special case of copy elision, which is described as (§12.8 [class.copy]/p31):

When certain criteria are met, an implementation is allowed to omit the copy/move construction of a class object, even if the constructor selected for the copy/move operation and/or the destructor for the object have side effects.

This allows implementations to elide "copy/move construction"; constructing an object using something that's neither a copy constructor nor a move constructor is not "copy/move construction".

Because C has a user-defined destructor, an implicit move constructor is not generated. Thus, overload resolution will select the templated constructor with Args deduced as C, which is a better match than the implicit copy constructor for rvalues. However, the compiler can't elide calls to this constructor, as it has side effects and is neither a copy constructor nor a move constructor.

If the templated constructor is instead

template<class ...Args>
C(Args ... args) {std::cout << "Ctr\n";} 

Then it can't be instantiated with Args = C to produce a copy constructor, as that would lead to infinite recursion. There's a special rule in the standard prohibiting such constructors and instantiations (§12.8 [class.copy]/p6):

A declaration of a constructor for a class X is ill-formed if its first parameter is of type (optionally cv-qualified) X and either there are no other parameters or else all other parameters have default arguments. A member function template is never instantiated to produce such a constructor signature.

Thus, in that case, the only viable constructor would be the implicitly defined copy constructor, and calls to that constructor can be elided.

If we instead remove the custom destructor from C, and add another class to track when C's destructor is called instead:

struct D {
    ~D() { std::cout << "D's Dstr\n"; }
};

template<class ...ArgsIn>
struct C {
  template<class ...Args>
  C(Args&& ... args) {std::cout << "Ctr\n";}
  D d;
};

We see only one call to D's destructor, indicating that only one C object is constructed. Here C's move constructor is implicitly generated and selected by overload resolution, and you see RVO kick in again.

来源：https://stackoverflow.com/questions/24925137/c-universal-reference-in-constructor-and-return-value-optimization-rvo