我正在尝试为类型列表的可变参数模板版本（受Modern C++ Design启发）实现 ReplaceAll 元函数，到目前为止，我可以使用以下代码实现所需的结果：

template <typename... Elements> struct Typelist {};

//
// helper metafunction to set "Head" of typelist
//
template <typename TList, typename T> struct PushFrontT;

template <typename... Elements, typename T>
struct PushFrontT<Typelist<Elements...>, T> {
    using Result = Typelist<T, Elements...>;
};

//
// metafunction to replace all occurences of "T" with "U"
//
template <typename TList, typename T, typename U> struct ReplaceAll;

template <typename T, typename U>
struct ReplaceAll<Typelist<>, T, U> {
    using Result = Typelist<>;
};

template <typename... Tail, typename T, typename U>
struct ReplaceAll<Typelist<T, Tail...>, T, U> {
    using Result = typename PushFrontT<
        typename ReplaceAll<Typelist<Tail...>, T, U>::Result, U
    >::Result;
};

template <typename Head, typename... Tail, typename T, typename U>
struct ReplaceAll<Typelist<Head, Tail...>, T, U> {
    using Result = typename PushFrontT<
        typename ReplaceAll<Typelist<Tail...>, T, U>::Result, Head
    >::Result;
};

它以Typelist<T1, T2, T3>( 有效地用T目标 type替换所有出现的type U)的形式返回一个类型列表。

现在的问题是，当我尝试不使用 helper 元函数时PushFrontT，会以Typelist<T1, Typelist<T2, Typelist<T3, Typelist<>>>>不正确的形式创建嵌套的类型列表结构（尽管替换了Twith 的所有实例U）。

错误版本的代码如下：

template <typename T, typename U>
struct ReplaceAll<Typelist<>, T, U> {
    using Result = Typelist<>;
};

template <typename... Tail, typename T, typename U>
struct ReplaceAll<Typelist<T, Tail...>, T, U> {
    using Result = Typelist<U,
        typename ReplaceAll<Typelist<Tail...>, T, U>::Result
    >;
};

template <typename Head, typename... Tail, typename T, typename U>
struct ReplaceAll<Typelist<Head, Tail...>, T, U> {
    using Result = Typelist<Head,
        typename ReplaceAll<Typelist<Tail...>, T, U>::Result
    >;
};

基于我对可变参数模板的有限了解，我认为附加的Typelist是包扩展的副作用，但我不确定。
这是一个简单的测试程序来检查上述代码：

#include <type_traits>

int main () {
    using tlist = Typelist<int, char, int, double>;

    static_assert(
        std::is_same<
        typename ReplaceAll<tlist, int, long>::Result,
        Typelist<long, char, long, double>>::value,
        "Incorrect typelist!"
    );

    return(0);
}

简而言之，我的问题是如何在Typelist不使用外部辅助元函数（例如PushFrontT?

回答

template <typename TList, typename T, typename U> struct ReplaceAll;

template <typename ... Ts, typename T, typename U>
struct ReplaceAll<Typelist<Ts...>, T, U>
{
    using Result = Typelist<std::conditional_t<std::is_same_v<Ts, T>, U, Ts>...>;
};