c++ - 可以默认初始化具有已删除默认构造函数的类类型吗？

Question

根据 cppref 关于 value initialization 的说法

if T is a class type with no default constructor or with a user-provided or deleted default constructor, the object is default-initialized;

但由于该类类型删除了默认构造函数，对象如何被默认初始化？

据我所知，类类型的默认初始化需要默认构造函数的访问。如果我们有:

struct A {
    A() = delete;
    int k;
};

然后 A *a = new A; 会失败，A* a = new A(); 也会失败。

但是 A a{}; 没问题。但为什么？根据cppreference

Otherwise, If the braced-init-list is empty and T is a class type with a default constructor, value-initialization is performed.

Answer 1

我认为标准只是意味着“如果 T 是一个类类型，删除了默认构造函数，然后转到默认初始化”。它最终会失败，因为选择用于默认初始化的构造函数已被删除。它用于区分第二种情况，即“如果 T 是一个类类型，其默认构造函数既不是用户提供的也不是删除的”，对于这种情况，首先执行零初始化，然后执行默认- 如果 T 具有非平凡的默认构造函数，则初始化。

A a{} is OK,but why?

因为 A 是 aggregate type执行聚合初始化。请注意，自 C++11 起，聚合类型允许显式删除构造函数。

In all cases, if the empty pair of braces {} is used and T is an aggregate type, aggregate-initialization is performed instead of value-initialization.

和

An aggregate is one of the following types:

• ...
• class type (typically, struct or union), that has
  • ...
  • no user-provided, inherited, or explicit constructors (explicitly defaulted or deleted constructors are allowed) (since C++17) (until C++20)
  • ...

编辑

从 C++20 开始，行为发生了变化； A a{}; 会失败。

An aggregate is one of the following types:

• ...
• class type (typically, struct or union), that has
  • ...
  • no user-declared or inherited constructors (since C++20)
  • ...

A 不再是聚合。 A a{}; 执行 value-initialization ( default-initialization )，删除的构造函数被使用并失败。