大括号的类型如何影响 C++ 中的对象生命周期？

Question

我的一个朋友给我看了一个 C++20 的程序：

#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";
}

在 GCC 中它打印：

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

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

因此在情况 y 和 z 中 A 的生命周期延长了。

在Visual Studio中结果不同：

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

所以A的生命周期只有在y的情况下才会延长。

能否请您解释一下为什么大括号的类型会影响对象的生命周期？

Answer 1

Gcc 是正确的。只有在聚合初始化中使用列表初始化语法（即使用大括号）时，临时对象的生命周期才会 extended。

(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

对于direct initialization：

（强调我的）

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)

大括号的类型如何影响 C++ 中的对象生命周期？

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

c++

initialization

object-lifetime

temporary-objects

c++20