移动派生类的构造函数

问题描述

我有2个课程:

template<typename T>
class base{
    T t;
public:
    base(base &&b): t(std::move(b.t)){}
};

template<typename T, typename T2>
class derived : protected base<T>{
    T2 t2;
public:
    derived(derived &&d): base<T>(std::move(d)), t2(std::move(d.t2)){}
};

我将整个d对象移到derived move-constructor中以初始化base部分，并且d变为无效，但是我仍然需要它来将其用于t2初始化

I move entire d object in the derived move-constructor to initialize base part and d becomes invalid but I still need it to use it's part for t2 initialization

有可能做这样的事情吗?

Is it possible to do such a thing?

推荐答案

我会说您的构造是正确的，除了一些语法错误，您需要在初始化列表中限定base<T>:

I would say that your construct is correct except for a little syntax error, you need to qualify base<T> in the initializer list:

derived(derived &&d): base<T>(std::move(d)), t2(std::move(d.t2)){}

首先，初始化的顺序与初始化列表的顺序无关. n4296草案在 12.6.2中声明初始化基础和成员[class.base.init]§13

First, the order of initialization is independant of the order of the initializer list. Draft n4296 says in 12.6.2 Initializing bases and members [class.base.init] § 13

[注意:声明顺序是强制性的，以确保在子对象中销毁基础和成员子对象. 初始化的相反顺序. —尾注]

[ Note: The declaration order is mandated to ensure that base and member subobjects are destroyed in the reverse order of initialization. —end note ]

我们也有§7或同一章说:

We have also §7 or same chapter that says:

我的理解是，标准表示在derived类的移动ctor中，事情按以下顺序发生:

My understanding is that standard says that in the move ctor for the derived class, things happens in that order:

• 为基类移动ctor称为
  • 依次调用T的move ctor有效地构造目标的t成员，并最终将源的t成员清零
  • move ctor for base class is called
    • in turn it calls move ctor for T effectively constructing t member of target and eventually zeroing t member of source

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