`= default` move构造函数是否等效于成员级move构造函数? [英] Is a `=default` move constructor equivalent to a member-wise move constructor?
问题描述
是这个
struct Example {
int a, b;
Example(int mA, int mB) : a{mA}, b{mB} { }
Example(const Example& mE) : a{mE.a}, b{mE.b} { }
Example(Example&& mE) : a{move(mE.a)}, b{move(mE.b)} { }
Example& operator=(const Example& mE) { a = mE.a; b = mE.b; return *this; }
Example& operator=(Example&& mE) { a = move(mE.a); b = move(mE.b); return *this; }
}
与此等效
struct Example {
int a, b;
Example(int mA, int mB) : a{mA}, b{mB} { }
Example(const Example& mE) = default;
Example(Example&& mE) = default;
Example& operator=(const Example& mE) = default;
Example& operator=(Example&& mE) = default;
}
?
推荐答案
是相同的.
但是
struct Example {
int a, b;
Example(int mA, int mB) : a{mA}, b{mB} { }
Example(const Example& mE) = default;
Example(Example&& mE) = default;
Example& operator=(const Example& mE) = default;
Example& operator=(Example&& mE) = default;
}
此版本允许您跳过主体定义.
This version will permits you to skip the body definition.
但是,在声明explicitly-defaulted-functions
时必须遵循一些规则:
However, you have to follow some rules when you declare explicitly-defaulted-functions
:
8.4.2显式默认的功能[dcl.fct.def.default]
以下形式的函数定义:
attribute-specifier-seqopt decl-specifier-seqopt declarator virt-specifier-seqopt = default ;
被称为显式默认的定义.明确默认的功能应
is called an explicitly-defaulted definition. A function that is explicitly defaulted shall
-
是特殊的成员函数,
be a special member function,
具有相同的声明函数类型(除了可能不同的 ref-qualifiers ,并且在复制构造函数或复制赋值运算符的情况下,参数类型可以是对非常量T
",其中T
是成员函数的类的名称),就好像它已隐式声明一样,
have the same declared function type (except for possibly differing ref-qualifiers and except that in the case of a copy constructor or copy assignment operator, the parameter type may be "reference to non-const T
", where T
is the name of the member function’s class) as if it had been implicitly declared,
没有默认参数.
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