如何使用 C++ 转发引用正确指定 noexcept？

Question

举个例子：

#include <iostream>
#include <type_traits>
#include <utility>


struct Bar
{
    Bar() = default;
    Bar(Bar const&) noexcept(false) = default;
    Bar(Bar&&) noexcept(true) = default;
};

struct Baz
{
    Baz() = default;
    Baz(Baz const&) noexcept(true) = default;
    Baz(Baz&&) noexcept(false) = default;
};

template<typename T>
class Foo
{
    template<typename U, typename V>
    using enable_if_same = std::enable_if<std::is_same<typename std::remove_reference<U>::type, V>::value, U>;

public:
    template<typename U>
    Foo(typename enable_if_same<U, T>::type&& val) // noexcept( ? )
        : _val(std::forward<U>(val))
    {
    }

protected:
    T _val;
};


int main(void)
{
    std::cout << "Is the Bar copy constructor noexcept? " << std::is_nothrow_copy_constructible<Bar>::value << "\n";
    std::cout << "Is the Bar move constructor noexcept? " << std::is_nothrow_move_constructible<Bar>::value << "\n";
    std::cout << "Is the Foo<Bar> copy constructor noexcept? " << std::is_nothrow_copy_constructible<Foo<Bar>>::value << "\n";
    std::cout << "Is the Foo<Bar> move constructor noexcept? " << std::is_nothrow_move_constructible<Foo<Bar>>::value << "\n";

    std::cout << "\n";

    std::cout << "Is the Baz copy constructor noexcept? " << std::is_nothrow_copy_constructible<Baz>::value << "\n";
    std::cout << "Is the Baz move constructor noexcept? " << std::is_nothrow_move_constructible<Baz>::value << "\n";
    std::cout << "Is the Foo<Baz> copy constructor noexcept? " << std::is_nothrow_copy_constructible<Foo<Baz>>::value << "\n";
    std::cout << "Is the Foo<Baz> move constructor noexcept? " << std::is_nothrow_move_constructible<Foo<Baz>>::value << "\n";

    return 0;
}

编译并运行以上代码产生预期的输出：

Is the Bar copy constructor noexcept? 0
Is the Bar move constructor noexcept? 1
Is the Foo<Bar> copy constructor noexcept? 0
Is the Foo<Bar> move constructor noexcept? 1

Is the Baz copy constructor noexcept? 1
Is the Baz move constructor noexcept? 0
Is the Foo<Baz> copy constructor noexcept? 1
Is the Foo<Baz> move constructor noexcept? 0

但是，我很想知道是否可以显式指定 Foo(U&&) 构造函数的 noexcept-ness，采用转发引用（即我需要替换什么 ? with 在上面注释掉的 noexcept 说明符中)?

Answer 1

你可以这样做：

template<typename U, typename = 
    std::enable_if_t<std::is_same_v<std::remove_reference_t<U>, T>>>
Foo(U&& val) noexcept(noexcept(T(std::forward<U>(val))))
    : _val(std::forward<U>(val))
{}

里面的noexcept是一个noexcept operator, the outer one is a noexcept specifier.

Demo

我不得不改变 SFINAE 的使用方式，因为在你的构造函数中

template<typename U>
Foo(typename enable_if_same<U, T>::type&& val) 
    : _val(std::forward<U>(val))
{}

无法推导出类型 U（即使可以，...::type&& 也不会成为转发引用）。

如何使用 C++ 转发引用正确指定 noexcept？

How to properly specify noexcept with a C++ forwarding reference?

c++

perfect-forwarding

noexcept

forwarding-reference