为什么在删除默认构造函数 A::A() 时 'A a{};' 可以编译？

Question

这里是有问题的代码示例：

struct A {
    A() = delete;
};

int main()
{
//  A a(); // compiles, since it's a function declaration (most vexing parse)
//  A a;   // does not compile, just as expected
    A a{}; // compiles, why? The default constructor is deleted.
}

尝试使用任何可用的编译器here。我尝试了几个，但没有找到一个给出编译错误的。

Answer 1

这是一个当前的语言问题，很可能很快就会得到解决。可以找到解决必要设计变更的提案 here。来自提案摘要：

C++ currently allows some types with user-declared constructors to be initialized via aggregate initialization, bypassing those constructors. The result is code that is surprising, confusing, and buggy

Answer 2

因为A是一个aggregate type，那么给定的A a{};进行聚合初始化。

Each direct public base, (since C++17) array element, or non-static class member, in order of array subscript/appearance in the class definition, is copy-initialized from the corresponding clause of the initializer list.

在聚合初始化中，每个成员或元素（如果有的话）将被直接复制初始化，构造函数被绕过；所以 deleteed 与否并不重要。

注意聚合类型允许显式删除构造函数 (C++11 起) (C++20 前)，

no user-provided constructors (explicitly defaulted or deleted constructors are allowed) (since C++11) (until C++17)

no user-provided, inherited, or explicit constructors (explicitly defaulted or deleted constructors are allowed) (since C++17) (until C++20)

为什么在删除默认构造函数 A::A() 时 'A a{};' 可以编译？

Why does 'A a{};' compile when the default constructor A::A() is deleted?

c++

language-lawyer

uniform-initialization

c++11