a =默认移动构造函数等于成员式移动构造函数? [英] Does a =default move constructor equals to a member-wise move constructor?
问题描述
这是
struct示例{
int a,b;
示例(int mA,int mB):a {mA},b {mB} {}
示例(const Example& mE):a {mE.a},b {mE.b} }
示例(示例&&&mE):a {move(mE.a)},b {move(mE.b)} {}
示例& operator =(const Example& mE){a = mE.a; b = mE.b; return * this; }
示例& operator =(Example&& mE){a = move(mE.a); b = move(mE.b); return * this; }
}
相当于
struct示例{
int a,b;
示例(int mA,int mB):a {mA},b {mB} {}
示例(const Example& mE)
示例(示例&& mE)= default;
示例& operator =(const Example& mE)= default;
示例& operator =(Example&& mE)= default;
}
?
但
$ b $
struct示例{
int a,b;
示例(int mA,int mB):a {mA},b {mB} {}
示例(const Example& mE)
示例(示例&& mE)= default;
示例& operator =(const Example& mE)= default;
示例& operator =(Example&& mE)= default;
}
此版本将允许您跳过正文定义。
$但是,当你声明显式默认函数
时,你必须遵循一些规则:
8.4.2显式默认函数[dcl.fct.def.default]
一个形式的函数定义:
属性指定声明声明符virt-speci- er-seqopt =
称为显式默认定义。明确默认的函数
是一个特殊的成员函数,
具有相同的声明函数类型(除了可能不同的 ref-qualifiers ,除非在复制构造函数或复制赋值运算符的情况下,参数类型可以是非常量
T
,其中T
是成员函数的类的名称) ,
Is this
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; }
}
equivalent to this
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;
}
?
Yes both are the same.
But
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;
}
This version will permits you to skip the body definition.
However, you have to follow some rules when you declare explicitly-defaulted-functions
:
8.4.2 Explicitly-defaulted functions [dcl.fct.def.default]
A function definition of the form:
attribute-specifier-seqopt decl-specifier-seqopt declarator virt-specifier-seqopt = default ;
is called an explicitly-defaulted definition. A function that is explicitly defaulted shall
be a special member function,
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
", whereT
is the name of the member function’s class) as if it had been implicitly declared,not have default arguments.
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