C++移动赋值触发器首先移动构造函数

Question

#include <iostream>    

class C {

public:

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

  C() { std::cout << this << " constructor\n"; }

  C(C&& rhs) {
    std::cout << &rhs << " rhs\n";
    std::cout << this << " move constructor\n";
  }

  C& operator=(C&& rhs) {
    std::cout << &rhs << " rhs\n";
    std::cout << this << " move assignment\n";
    return *this;
  }

};

C make_one() {
  C tmp;
  return tmp;
}

int main() {

  std::cout << "move constructor:\n";
  C c1(make_one());
  std::cout << &c1 << " &c1\n\n";

  std::cout << "move assignment:\n";
  C c2;
  c2 = make_one();
  ...
}

输出:

move constructor:
000000000021F9B4 constructor         // tmp constructed in make_one()
000000000021F9B4 rhs                 // return triggers tmp being passed to ...
000000000021FA04 move constructor    // ... c1's move constructor (see below)
000000000021F9B4 destructor          // tmp destructs on going out of scope
000000000021FA04 &c1                 // (confirmed c1's address)

move assignment:
000000000021FA24 constructor         // c2 constructed
000000000021F9B4 constructor         // tmp constructed in make_one() again
000000000021F9B4 rhs                 // tmp passed to ...
000000000021FA34 move constructor    // ... a new object's move constructor
000000000021F9B4 destructor          // tmp destructs on going out of scope
000000000021FA34 rhs                 // new object passed to ...
000000000021FA24 move assignment     // .. c2's move assignment operator
000000000021FA34 destructor          // new object destructs
...

移动分配似乎首先触发移动构造函数并创建一个额外的对象.这是正常的吗？我希望(通过类比复制赋值)将tmp直接传递给c2的移动赋值.

[Visual Studio Express 2013]

Answer 1

"额外对象"称为返回值.从函数返回值时; 此值是从您提供给return语句的值构造的复制/移动.

通常这会经历复制省略,这可以解释为什么你不认识它.当复制省略发生时,该行C tmp;实际上将tmp直接构造为返回值.复制省略也可以在其他一些情况下发生; 有关全文,请参阅C++ 11 [class.copy]#31.

大概你可以在这里手动禁用复制省略,或者编译器决定不执行复制省略是个好主意.更新:您的编译器仅在发布版本上执行此特定的copy-elision - 感谢Praetorian