c++ - `=default` move 构造函数是否等同于成员 move 构造函数？

Question

这是

struct Example { 
    string a, b; 

    Example(Example&& mE) : a{move(mE.a)}, b{move(mE.b)} { }
    Example& operator=(Example&& mE) { a = move(mE.a); b = move(mE.b); return *this; } 
}

相当于这个

struct Example { 
    string a, b;

    Example(Example&& mE)            = default;
    Example& operator=(Example&& mE) = default;
}

?

Answer 1

是的，两者都是一样的。

但是

struct Example { 
    string a, b; 

    Example(Example&& mE)            = default;
    Example& operator=(Example&& mE) = default;
}

此版本允许您跳过正文定义。

但是，当你声明 explicitly-defaulted-functions 时，你必须遵守一些规则:

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”, where T is the name of the member function’s class) as if it had been implicitly declared,

• not have default arguments.