How can the type of braces influence object lifetime in C++?

Question

A friend of mine showed me a program in C++20:

#include <iostream>

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

struct B
{
    const A &a;
};

int main()
{
    B x({});
    std::cout << "---\n";
    B y{{}};
    std::cout << "---\n";
    B z{A{}};
    std::cout << "---\n";
}

In GCC it prints:

A()
~A()
---
A()
---
A()
---
~A()
~A()

https://gcc.godbolt.org/z/ce3M3dPeo

So the lifetime of A is prolonged in cases y and z.

In Visual Studio the result is different:

A()
~A()
---
A()
---
A()
~A()
---
~A()

So the lifetime of A is only prolonged in case y.

Could you please explain why the type of braces influences the object lifetime?

Answer 1

Gcc is correct. The lifetime of the temporary will be extended only when using list-initialization syntax (i.e. using braces) in initialization of an aggregate.

(since C++20) a temporary bound to a reference in a reference element of an aggregate initialized using direct-initialization syntax (parentheses) as opposed to list-initialization syntax (braces) exists until the end of the full expression containing the initializer.

struct A {
  int&& r;
};
A a1{7}; // OK, lifetime is extended
A a2(7); // well-formed, but dangling reference

For direct initialization:

(emphasis mine)

otherwise, if the destination type is a (possibly cv-qualified) aggregate class, it is initialized as described in aggregate initialization except that narrowing conversions are permitted, designated initializers are not allowed, a temporary bound to a reference does not have its lifetime extended, there is no brace elision, and any elements without an initializer are value-initialized. (since C++20)