What is the lifetime of reference member default initializer in C++?

Question

Please consider this short code example:

#include <iostream>

struct A
{
    A() { std::cout << "A() "; }
    ~A() { std::cout << "~A() "; }
};

struct B { const A &a = A(); };

int main()
{
    B x;
    std::cout << ". ";
    auto y = B();
    std::cout << ". ";
    auto z = B{};
    std::cout << ". ";
}

GCC prints here ( https://gcc.godbolt.org/z/a83bn54qT ):

A() ~A() . A() ~A() . A() . ~A() 

meaning that the lifetime of A-object initializing reference in x and y is short, but in z the lifetime is prolonged till the end of the scope.

Could you please explain why it is so from C++ standard point of view?

Answer 1

Lifetime extension is not guaranteed since the code is ill-formed.

[class.base.init]/11:

A temporary expression bound to a reference member from a default member initializer is ill-formed. [Example 8:

struct A {
  A() = default;        // OK
  A(int v) : v(v) { }   // OK
  const int& v = 42;    // OK
};
A a1;                   // error: ill-formed binding of temporary to reference
A a2(1);                // OK, unfortunately

— end example]