移动构造函数是隐式的吗？

问题描述

考虑下面的类：

  class A 
 {
 public：
 std :: string field_a; 
 std :: string field_b; 
} 
  

现在考虑以下副本构造：

  A a1（a2）; 
  

复制结构将充分复制 A 缺少一个显式的拷贝构造函数，因为 std :: string 的拷贝构造函数将被编译器生成的隐式拷贝构造函数调用。

$

编辑：测试显示：

  A a2（std :: move（a1））; 
  

实际上会导致复制构造，除非特定的move构造函数：

  A（A&& other）：a（std :: move（other.a））{} 
  
 
 
 
已定义。
 
 
 
  EDIT EDIT  
我ping了Stephan T Lavavej，问他为什么VC 2012似乎不遵循12.8中关于隐式移动构造函数生成的草案。他很善良地解释：
 
 
解决方案
是的，从C ++ 11草案，12.8：
后面的条件指定更多细节：
很明显，移动构造函数将被隐式声明为：
 
 
该类没有用户声明的任何其他特殊成员函数。
 
 
移动构造函数可以通过移动其所有成员和基数来实现。
 
 
您的类显然符合这些条件。
 
Consider the following class:
class A
{
public:
   std::string field_a;
   std::string field_b;
}
Now consider the following copy construction:
A a1(a2);
The copy construction will adequately copy A despite the lack of of an explicit copy constructor because the copy constructors for std::string will be called by the compiler generated implicit copy constructor.
What I wish to know is, is the same true for move construction?
EDIT: Testing here shows that:
A a2(std::move(a1));
Will actually result in a copy construction, unless the specific move constructor:
A( A && other ) : a(std::move(other.a)) {}
Is defined.
EDIT EDITI pinged Stephan T Lavavej and asked him why VC 2012 doesn't seem to follow what draft 12.8 states regarding implicit move constructor generation.  He was kind enough to explain:
 解决方案 
Yes, from the C++11 draft, 12.8:
The latter condition is specified with more detail later:
Plainly speaking, the move constructor will be implicitly declared if:
The class does not have user-declared any of the other special member functions.
The move constructor can be sensibly implemented by moving all its members and bases.
Your class obviously complies with these conditions.
                        
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