如何在任何嵌套级别替换模板中的类型？

Question

我有一个名为 的类型the_bad。the_good我想用模板元函数替换该类型。我预计该模板元函数会非常复杂，因此我们将其称为the_ugly.

我的问题是它the_bad可以嵌套在其他模板中，并且我想保持其他模板和参数不变。这是我的意思的一个例子，其中的实现不完整the_ugly：

struct the_good {};
struct the_bad {};

template<typename T>
struct the_ugly_impl {
    using type = T;
};

template<>
struct the_ugly_impl<the_bad> {
    using type = the_good;
};

template<typename T>
using the_ugly = typename the_ugly_impl<T>::type;

// passes, yay!
static_assert(std::same_as<the_good, the_ugly<the_bad>>);

// doesn't pass
static_assert(std::same_as<std::vector<the_good>, the_ugly<std::vector<the_bad>>>);

// doesn't pass
static_assert(std::same_as<std::list<std::vector<the_good>>, the_ugly<std::list<std::vector<the_bad>>>>);

我该如何修复，the_ugly以便它在任何嵌套级别替换the_bad为，同时保持所有其他模板和其他模板参数不变？the_good

Answer 1

You just need to handle the template case, and have that invoke itself recursively:

template<typename T>
struct the_ugly_impl {
    using type = T;
};

template<typename T>
using the_ugly = typename the_ugly_impl<T>::type;

template<>
struct the_ugly_impl<the_bad> {
    using type = the_good;
};

template <template <typename...> class L, typename... Ts>
struct the_ugly_impl<L<Ts...>> {
    using type = L<the_ugly<Ts>...>;
};