c++ - 为什么 `std:variant` 的 `operator=(T&& t)` 的 noexcept 规范不依赖于内部类型的析构函数的 noexcept 规范？

Question

长标题:为什么 std:variant 的 operator=(T&& t) 的 noexcept 规范不依赖于内部类型的析构函数的 noexcept 规范？

我可以在 cppreference 上看到那个

template <class T> variant& operator=(T&& t) noexcept(/* see below */);

是

noexcept(std::is_nothrow_assignable_v<T_j&, T> && 
std::is_nothrow_constructible_v<T_j, T>)

所以编译:

struct FooThrow {
  ~FooThrow() noexcept(false) {throw;} 
};
static_assert(std::is_nothrow_assignable_v<std::variant<FooThrow, int>, int>);

但是它调用了 FooThrow 的析构函数，它是 noexcept(false):

std::variant<FooThrow, int> x;
x = 3; // throws

好像不太对。我错过了什么吗？

Answer 1

一般来说，标准库类型不会友好地接受具有抛出析构函数的类型。或者具体来说，当析构函数实际发出异常时。有一个关于它的一般规则 ( [res.on.functions] )

In certain cases (replacement functions, handler functions, operations on types used to instantiate standard library template components), the C ++ standard library depends on components supplied by a C ++ program. If these components do not meet their requirements, this International Standard places no requirements on the implementation.

In particular, the effects are undefined in the following cases:

...

• if any replacement function or handler function or destructor operation exits via an exception, unless specifically allowed in the applicable Required behavior: paragraph.

因为 variant::operator= 没有关于抛出析构函数的特殊声明，所以让那些析构函数实际抛出是 UB。