class 删除的析构函数被认为是可简单复制的？

Question

这可以用 gcc 和 clang 编译

#include <type_traits>

struct A {
    ~A() = delete;
};

static_assert(std::is_trivially_copyable_v<A>);
int main() { }

删除了析构函数的class是否可以简单复制？

Answer 1

我假设A的析构函数被隐式删除，= delete;没有影响，不计入std::is_trivially_copyable_v<A>。

扩展示例也是如此

#include <type_traits>

struct A {
    int a;
    int b;
    ~A() = delete;
};

static_assert(std::is_trivially_copyable_v<A>);
int main() { }

The implicitly-declared or defaulted destructor for class T is defined as deleted (since C++11) if any of the following is true:

has a non-static data member that cannot be destructed (has deleted or inaccessible destructor)

...

= delete 不会改变析构函数的属性“已删除隐式声明的析构函数”，因此上面的 class A 与下面的 class A 相同：

struct A {
    int a;
    int b;
};

条件“有一个普通的未删除的析构函数”为真。

我只有一个想法。由于隐式析构函数已经不可用，因此 具有一个普通的未删除析构函数 的要求不会被不可用析构函数的 = delete 违反，无需删除。但是 = delete 使用户声明的隐式删除析构函数（使其可用）。让我们看看 int main() { A a; } 编译器不会像您的 A 示例那样调用不可用的析构函数，而是必须像我的 A 示例那样调用用户声明的析构函数，并且它被删除（被可用）。

Answer 2

可简单复制的 class 可能没有已删除的构造函数

考虑到该主题的标题和未来的读者：标准完全清楚 A 定义为

struct A {
    ~A() = delete;
};

根据 [class.prop]/1，

不是简单可复制的 class，特别是 /1.3

A trivially copyable class is a class:

(1.1) that has at least one eligible copy constructor, move constructor, copy assignment operator, or move assignment operator ([special], [class.copy.ctor], [class.copy.assign]),

(1.2) where each eligible copy constructor, move constructor, copy assignment operator, and move assignment operator is trivial, and

(1.3) that has a trivial, non-deleted destructor ([class.dtor]).

至于为什么 GCC 和 Clang 都不遵守这一点，在 Clang 错误报告中

Bug 39050 - is_trivially_copyable_v misbehaves

理查德史密斯评论

(R. Smith) This is DR1734, which Clang (and apparently GCC) does not yet implement.

之后讨论了为什么这很难实现（ABI 兼容性问题），但没有提及实际尝试提交 DR 或类似文件以更改或挑战标准文本；例如：

(R. Smith) [...] The part of the ABI I was referring to is "POD for the purpose of layout", for which various different targets use different rules, and the four that I listed above base their rule on Clang's broken notion of "trivially-copyable".

因此，从语言律师的角度来看，该主题的标准非常明确：A 不可简单复制。

class 删除的析构函数被认为是可简单复制的？

class with deleted destructor considered trivially copyable?

c++

gcc

clang

language-lawyer

可简单复制的 class 可能没有已删除的构造函数