c++ - 删除析构函数的类被认为是可简单复制的？

Question

这用 gcc 和 clang 编译

#include <type_traits>

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

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

删除析构函数的类是否可以简单地复制？

Answer 1

可简单复制的类可能没有删除的构造函数
鉴于该主题的标题和 future 的读者:该标准完全明确 A定义为

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

根据 [class.prop]/1，不是可简单复制的类, 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 都不尊重这一点，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.

之后讨论继续讨论为什么这很难实现(ABI 兼容性问题)，但没有提及实际尝试提交 DR 或类似内容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是 不是 可复制。