C++ 自动生成默认/复制/移动构造函数和复制/移动赋值运算符的条件？
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Conditions for automatic generation of default/copy/move ctor and copy/move assignment operator?
提问by oompahloompah
I want to refresh my memory on the conditions under which a compiler typically auto generates a default constructor, copy constructor and assignment operator.
我想刷新我对编译器通常自动生成默认构造函数、复制构造函数和赋值运算符的条件的记忆。
I recollect there were some rules, but I don't remember, and also can't find a reputable resource online. Can anyone help?
我记得有一些规则，但我不记得了，而且在网上也找不到信誉良好的资源。任何人都可以帮忙吗？
回答by Philipp
In the following, "auto-generated" means "implicitly declared as defaulted, but not defined as deleted". There are situations where the special member functions are declared, but defined as deleted.
在下文中，“自动生成”表示“隐式声明为默认值，但未定义为已删除”。在某些情况下，特殊成员函数已声明，但定义为已删除。
- The default constructor is auto-generated if there is no user-declared constructor (§12.1/5).
- The copy constructor is auto-generated if there is no user-declared move constructor or move assignment operator (because there are no move constructors or move assignment operators in C++03, this simplifies to "always" in C++03) (§12.8/8).
- The copy assignment operator is auto-generated if there is no user-declared move constructor or move assignment operator (§12.8/19).
- The destructor is auto-generated if there is no user-declared destructor (§12.4/4).
- 如果没有用户声明的构造函数（第 12.1/5 节），则默认构造函数是自动生成的。
- 如果没有用户声明的移动构造函数或移动赋值运算符，复制构造函数将自动生成（因为在 C++03 中没有移动构造函数或移动赋值运算符，这在 C++03 中简化为“始终”）（第 12.8/8 节）。
- 如果没有用户声明的移动构造函数或移动赋值运算符（第 12.8/19 节），则会自动生成复制赋值运算符。
- 如果没有用户声明的析构函数（第 12.4/4 节），析构函数将自动生成。
C++11 and later only:
仅限 C++11 及更高版本：
- The move constructor is auto-generated if there is no user-declared copy constructor, copy assignment operator or destructor, and if the generated move constructor is valid (§12.8/10).
- The move assignment operator is auto-generated if there is no user-declared copy constructor, copy assignment operator or destructor, and if the generated move assignment operator is valid (e.g. if it wouldn't need to assign constant members) (§12.8/21).
- 如果没有用户声明的复制构造函数、复制赋值运算符或析构函数，并且生成的移动构造函数有效（第 12.8/10 节），则移动构造函数是自动生成的。
- 如果没有用户声明的复制构造函数、复制赋值运算符或析构函数，并且生成的移动赋值运算符有效（例如，如果它不需要分配常量成员）（第 12.8 节/ 21)。
回答by Marco M.
I've found the diagram below very useful.
我发现下面的图表非常有用。
