C ++ 11移动构造函数

问题描述

考虑以下类别，实现移动构造函数的正确方法是：

  class C {
 public：
 C（）; 
 C（C& c）; 
 private：
 std :: string string; 
} 
  

当然，想法是避免复制 string 或重新分配它两次。

假设基本示例只是为了清楚起见，我需要一个移动构造函数。

holar我试过：

  C :: C（C& C）{
 // move ctor 
 string = std :: move（c.string）; 
} 
  

和

  C :: C（C& c）：string（std :: move（c.string））{
 // move ctor 
} 
  

在gcc 4.8上编译并运行良好。这似乎选项A是正确的行为， string 被复制，而不是移动选项B.Æ
这是 std :: string 正确实现移动构造函数

解决方案

它本身有一个move-ctor， C 的隐式定义的move-ctor将负责正确的移动操作。你不能自己定义它。但是，如果您有任何其他数据成员，并且具体：

您可能想实现自己的move-ctor。

在需要move-ctor的情况下，更喜欢初始化列表语法。总是！否则，你可能会得到一个默认的构造，每个对象在初始化器列表中没有提到（这是你只能为非默认ctors的成员对象）。

What would be the correct way to implement a move constructor considering the following class:

class C {
public:
    C();
    C(C&& c);
private:
    std::string string;
}

Of course, the idea is to avoid copying string or deallocating it twice.
Lets assume the basic example is just for clarity and I do need a move constructor.

I tried:

C::C(C&& c) {
    //move ctor
    string = std::move(c.string);
}

And

C::C(C&& c) : string(std::move(c.string)) {
    //move ctor
}

Both compile fine on gcc 4.8 and run fine. It seems option A is the correct behaviour, string gets copied instead of moved with option B.
Is this the correct implementation of a move constructor?

解决方案

Since std::string itself has a move-ctor, the implicitly defined move-ctor for C will take care of the proper move operation. You may not define it yourself. However, if you have any other data member and specifically:

you may want to implement your own move-ctor.

In case you need the move-ctor, prefer the initializer list syntax. Always! Otherwise, you may end up with a default construction per object not mentioned in the initializer list (which is what you're forced for member objects with non-default ctors only).

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