std::variant 和模棱两可的初始化

Question

考虑以下代码：

void fnc(int)
{
    std::cout << "int";
}

void fnc(long double)
{
    std::cout << "long double";
}

int main()
{
    fnc(42.3); // error
}

由于对 fnc.
的模糊调用，它给出了一个错误但是，如果我们写下一段代码：

std::variant<int, long double> v{42.3};
std::cout << v.index();

输出为1，表明选择了double->long double转换。据我所知，std::variant 遵循关于转换等级的 C++ 规则，但这个例子显示了不同之处。对这种行为有什么解释吗？

Answer 1

在P0608, variant<int, long double> v{42.3} also has the ambiguous issue之前，因为42.3可以转换为int或long double。

P0608 已更改 the behavior of variant's constructors:

template<class T> constexpr variant(T&& t) noexcept(see below);

Let Tj be a type that is determined as follows: build an imaginary function FUN(Ti) for each alternative type Ti for which Ti x[] = {std::forward<T>(t)}; is well-formed for some invented variable x and, if Ti is cv bool, remove_cvref_t<T> is bool. The overload FUN(Ti) selected by overload resolution for the expression FUN(std::forward<T>(t)) defines the alternative Tj which is the type of the contained value after construction.

在您的示例中，variant 有两种替代类型：int 和 long double，因此我们可以构建以下表达式

        int x[] = {std::forward<double>(42.3)}; // #1
long double y[] = {std::forward<double>(42.3)}; // #2

由于only #2 is well-formed，variant成功推导出包含的值类型long double。

std::variant 和模棱两可的初始化

std::variant and ambiguous initialization

c++

std

language-lawyer

c++17