Is a `=default` move constructor equivalent to a member-wise move constructor?

Question

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;
}

?

Answer 1

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

• not have default arguments.