"temporary of type 'A' has protected destructor"，但它的类型是B

Question

在以下代码中，使用 Clang 8.0.0+ 和 -std=c++17 编译，使用 B{} 创建派生的 class 实例会出现错误 error: temporary of type 'A' has protected destructor。当临时对象的类型为 B（因此应该有一个 public 析构函数）时，为什么 A 出现在此消息中？

https://godbolt.org/z/uOzwYa

class A {
protected:
    A() = default;
    ~A() = default;
};

class B : public A {
// can also omit these 3 lines with the same result
public:
    B() = default;
    ~B() = default;
};

void foo(const B&) {}

int main() {

    // error: temporary of type 'A' has protected destructor
    foo(B{});
    //    ^

    return 0;
}

Answer 1

这是 C++20 之前 aggregate initialization 的一个微妙问题。

在 C++20 之前，B（和 A）是 aggregate types:

（强调我的）

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

然后

If the number of initializer clauses is less than the number of members and bases (since C++17) or initializer list is completely empty, the remaining members and bases (since C++17) are initialized by their default member initializers, if provided in the class definition, and otherwise (since C++14) by empty lists, in accordance with the usual list-initialization rules (which performs value-initialization for non-class types and non-aggregate classes with default constructors, and aggregate initialization for aggregates).

所以B{}通过聚合初始化构造一个临时对象，它将直接用空列表初始化基础子对象，即进行聚合初始化构造A基础子对象。请注意， B 的构造函数被绕过了。问题是在这种情况下，无法调用 protected 析构函数来销毁直接构造的 A 类型的基类子对象。（它不会抱怨 protected 构造函数，因为它也被 A 的聚合初始化绕过了。）

可以改为foo(B());，避免聚合初始化； B()执行value-initialization，临时对象会被B的构造函数初始化，那就没问题了。

顺便说一句，自 C++20 以来，您的代码可以正常工作。

no user-declared or inherited constructors (since C++20)

B（和A）不再是聚合类型。 B{}执行list initialization，然后临时对象被B的构造函数初始化；效果与 B().

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