c++ - g++-4.8.1 认为显式声明的没有异常规范的析构函数总是 noexcept(true)

Question

考虑以下程序:

#include <type_traits>

struct Thrower
{
    ~Thrower() noexcept(false) { throw 1; }
};

struct Implicit
{
    Thrower t;
};
static_assert(!std::is_nothrow_destructible<Implicit>::value, "Implicit");

struct Explicit
{
    ~Explicit() {}

    Thrower t;
};
static_assert(!std::is_nothrow_destructible<Explicit>::value, "Explicit");

使用g++-4.8.1，Explicit上出现静态断言失败——似乎认为~Explicit()是 noexcept。这不符合我的期望。根据§12.4.3:

A declaration of a destructor that does not have an exception-specification is implicitly considered to have the same exception-specification as an implicit declaration

这里有趣的是 Implicit 的检查似乎是根据我对 §15.4.14 的解释(通过 §12.4.7)进行的。

...If f is an...destructor...it's implicit exception-specification specifies...f has the exception-specification noexcept(true) if every function it directly invokes allows no exceptions.

g++-4.7 缺少 is_nothrow_destructable，我编写了自己的代码来检查 4.7 中的行为。该程序似乎编译得很好。我保留完全错误的权利和我困惑的根源:

template <typename T>
struct is_nothrow_destructible
{
    static constexpr bool value = noexcept(std::declval<T>().~T());
};

TL;DR:为什么g++-4.8.1认为显式声明的没有异常规范的析构函数总是 noexcept(true)?

---

更新:我打开了一个错误:57645 .如果你真的需要解决这个问题，你可以在析构函数中添加一个异常规范(就像示例中的 Thrower 一样)。

Answer 1

TL;DR: Why does g++-4.8.1 think that an explicitly-declared destructor with no exception specification is always noexcept(true)?

因为它有错误？

您对标准的解释是正确的，并且 Clang 正确地实现了它(断言不会触发)。

f has the exception-specification noexcept(true) if every function it directly invokes allows no exceptions.

析构函数直接调用所有子对象的析构函数:

§12.4 [class.dtor] p8:

After executing the body of the destructor and destroying any automatic objects allocated within the body, a destructor for class X calls the destructors for X’s direct non-variant non-static data members, [...].