【发布时间】:2021-04-29 10:11:30
【问题描述】:
所以我一直在看这个:https://akrzemi1.wordpress.com/2017/12/02/your-own-type-predicate/
它表明您可以采用自定义特征(或谓词)is_thing<T>::type 并将其别名为 is_thing_t<T> - 例如:
template <bool Cond>
using enable_if_t = typename enable_if<Cond>::type;
现在我想用同样的技巧来实现其他特征中使用的值别名。我在下面尝试了一个 sn-p 代码:
// Primary template
// tparam 1 (T) is to pass in the class-under-test.
// tparam 2 (deafult to void) is to catch all cases not specialised - so they evaluate to false
template <typename T, typename = void>
struct is_really_bob : std::false_type {};
// specialisation
// tparam 1 (T) is to pass in the class-under-test.
// tparam 2 special meta-function that - if our class best-matches THIS template then it will evaluate to true
template <typename T>
struct is_really_bob <T, std::void_t<
typename T::result_type, // Has a nested type called result_type
decltype(T::check_bob()), // Has a static member fn called check_bob
decltype(std::is_default_constructible_v<T>), // required for below:
decltype(T{}.set_bob(std::string{""})), // has member fn set_bob that takes a string (also requires default c'tor)
decltype(T{}.get_bob()) // has member fn get_bob (also requires default c'tor)
>> : std::true_type {};
///// THIS FAILS /////
template <bool T>
using is_really_bob_v = typename is_really_bob<T>::value;
这里的问题是::value 不是类型名称。我不知道生成值别名的语法。有没有办法做到这一点?我希望最终能够做类似的事情:
if constexpr (is_really_bob_v<T>)
{
// ...
}
完整的代码示例在这里:https://godbolt.org/z/dG11vxesW - 它没有将 ::value 别名的尝试注释掉,因此我们至少可以看到其余代码正常工作
【问题讨论】:
标签: c++ templates template-meta-programming typetraits