c++ - 默认情况下是否继承构造函数 noexcept(true)？

Question

Here我发现:

Inheriting constructors [...] are all noexcept(true) by default, unless they are required to call a function that is noexcept(false), in which case these functions are noexcept(false).

这是否意味着在下面的示例中继承的构造函数是noexcept(true)，即使它已在基类中显式定义为noexcept(false) ，或者它本身被认为是一个不被调用的函数？

struct Base {
    Base() noexcept(false) { }
};

struct Derived: public Base {
    using Base::Base;
};

int main() {
    Derived d;
}

Answer 1

继承的构造函数也将是 noexcept(false) 因为正如你引用的那样，继承的构造函数默认情况下将是 noexcept(true)

unless they are required to call a function that is noexcept(false)

当 Derived 构造函数运行时，它还会调用 Base 构造函数，它是 noexcept(false)，因此，Derived 构造函数也将是 noexcept(false)。

以下证明了这一点。

#include <iostream>

struct Base {
  Base() noexcept(false) { }
};

struct Derived: public Base {
  using Base::Base;
};

int main() {
  std::cout << noexcept(Derived());
}

输出 0。