我理解您的问题是希望实现一种通用方法来测试给定类型是否为给定模板模板类型的特化。可以使用类模板进行操作,就像在
Frank的回答中所示。我将向您展示一种补充专业化的方法 - 使用函数重载的模板类型别名:
#include <type_traits>
#include <complex>
#include <iostream>
template <template <class...> class TT, class... Args>
std::true_type is_tt_impl(TT<Args...>);
template <template <class...> class TT>
std::false_type is_tt_impl(...);
template <template <class...> class TT, class T>
using is_tt = decltype(is_tt_impl<TT>(std::declval<typename std::decay<T>::type>()));
int main() {
static_assert(is_tt<std::complex, std::complex<int>>::value, "!");
static_assert(is_tt<std::complex, std::complex<float>>::value, "!");
static_assert(!is_tt<std::complex, float>::value, "!");
}
【演示】
您可以按照以下方式使用trait:
#include <type_traits>
#include <complex>
#include <iostream>
template <template <class...> class TT, class... Args>
std::true_type is_tt_impl(TT<Args...>);
template <template <class...> class TT>
std::false_type is_tt_impl(...);
template <template <class...> class TT, class T>
using is_tt = decltype(is_tt_impl<TT>(std::declval<typename std::decay<T>::type>()));
template <class T>
typename std::enable_if<is_tt<std::complex, T>::value>::type print(T t) {
std::cout << "(" << t.real() << "," << t.imag() << ")" << std::endl;
}
template <class T>
typename std::enable_if<!is_tt<std::complex, T>::value>::type print(T t) {
std::cout << t << std::endl;
}
int main() {
print(std::complex<int>(1, 2));
print(std::complex<double>(1.5, 2.5));
print(5.5);
}
(1,2)
(1.5,2.5)
5.5
[live demo]