默认的默认构造函数，为什么它不是用户提供的默认构造函数？

Question

For example, clang does not compile this code, because, the defaulted default constructor for struct A below, A() = default; is not considered to be user-provided.

struct A{ A() = default; };
const A a;

But if you look at [dcl.fct.def.general]/1 you'll see:

function-body:
ctor-initializer _opt compound-statement
function-try-block
= default ;
= delete ;

That is, = default; is the function body for the default constructor A::A() , which is the same as saying that the definition A() = default; above is equivalent to A(){} as {} is the body for a default constructor.

By the way, g++ compiles the snippet above, but I know g++ has other issues in this regard, according to this comment by Jonathan Wakely.

Answer 1

Because the standard says so ( [dcl.fct.def.default]/5 ):

A function is user-provided if it is user-declared and not explicitly defaulted or deleted on its first declaration.

Doing it this way allows you to maintain the triviality property with = default; . Otherwise, there's no way to give a class with another constructor a trivial default constructor.