Class 具有默认的受保护析构函数，但派生的 class 是不可破坏的？

Question

在下面的例子中，第一个静态断言被触发但第二个没有：

#include<type_traits>
struct A{
 protected:
  ~A()=default;
};
struct B:A{
  //this static assertion fails
  static_assert(std::is_trivially_destructible<A>::value,"");
};
//this static assertion succeeds
static_assert(std::is_trivially_destructible<B>::value,"");

（使用 GCC、Clang、MSVC、ellcc 进行检查）

我不明白为什么 A 不能 trivially_destructible，在 B 里面，而 B 是微不足道的可破坏的。这似乎与 C++ 标准的这两段矛盾，其中没有提到可访问性：

[class.dtor]

A destructor is trivial if it is not user-provided and if:

(6.1) — the destructor is not virtual,

(6.2) — all of the direct base classes of its class have trivial destructors, and

(6.3) — for all of the non-static data members of its class that are of class type (or array thereof), each such class has a trivial destructor.

[dcl.fct.def.default]

A function is user-provided if it is user-declared and not explicitly defaulted or deleted on its first declaration.

Answer 1

简单的说，因为从外部来看，A是不可破坏的！析构函数是protected，所以如果你有a A* ptr，调用delete ptr会编译失败

Answer 2

补充 Sebastian Redl 的回答：std::is_trivially_destructible<T>::value==false 并不意味着类型是 not trivially-destructible.

所以如果在模板编程中使用它，最好不要使用标准库类型特征，而是直接使用编译器内在。例如 gcc:

#include<type_traits>
struct A{
    protected:
    ~A(){};
};
struct B:A{
  //Bad: fails while the fact that A is trivially destructible
  //could be used to implement optimized function member in B.
  static_assert(std::is_trivially_destructible<A>::value,"");
  //Good: __has_trivial_destructor actualy return what is name says!!
  static_assert(__has_trivial_destructor(A),"");
};

Class 具有默认的受保护析构函数，但派生的 class 是不可破坏的？

Class with defaulted protected destructor not trivially-destructible but derived class is?

c++

c++11

language-lawyer

destructor