是大括号可构造的类型特征
如何检查特定类型 typename T 是否可从参数 typename ...Args 以 T{Args...} 的方式构造代码>?我知道 std::is_constructible<T, Args... > 来自 <type_traits> 的类型特征，但它使用括号，而不是花括号.我在编写类型特征方面没有太多经验，因此无法提供初始示例.作为简化，我们可以接受任何合理的断言，即使这不会导致一般性的太大损失.
How can I check whether specific type typename T is constructible from arguments typename ...Args in the manner T{Args...}? I aware of std::is_constructible< T, Args... > type trait from <type_traits>, but it works with parentheses, not curly braces. I do not have too much experience in writing of type traits, so I cannot provide initial example. As simplification we can accept any reasonable assertions, even if this leads to not too significant loss of generality.
推荐答案
template<class T, typename... Args>
decltype(void(T{std::declval<Args>()...}), std::true_type())
test(int);
template<class T, typename... Args>
std::false_type
test(...);
template<class T, typename... Args>
struct is_braces_constructible : decltype(test<T, Args...>(0))
{
};
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