Variadic别名模板到非可变参数类模板

时间:2015-08-01 23:26:04

标签: c++ templates c++11 variadic-templates

在尝试编写一个用于讨论元函数类的简单示例时,我写了以下内容:

#include <type_traits>

struct first {
    template <typename T, typename U>
    using apply = T;
};

template <typename C, typename... Args>
struct curry {
    struct type {
        template <typename... OtherArgs>
        using apply = typename C::template apply<Args..., OtherArgs...>;
    };
};

int main() {
    static_assert(std::is_same<first::apply<int, char>, int>::value, ""); // OK

    using AlwaysInt = curry<first, int>::type;
    static_assert(std::is_same<AlwaysInt::apply<char>, int>::value, ""); // error
}

第二个static_assert无法在gcc 5.1上编译:

main.cpp:17:72: error: pack expansion argument for non-pack parameter 'U' of alias template 'template<class T, class U> using apply = T'
         using apply = typename C::template apply<Args..., OtherArgs...>;
                                                                        ^

和clang 3.6:

main.cpp:17:59: error: pack expansion used as argument for non-pack parameter of alias template
        using apply = typename C::template apply<Args..., OtherArgs...>;
                                                          ^~~~~~~~~~~~

两种情况都有相同的错误。但是,如果我将curry中的应用程序外包给单独的元函数:

template <typename C, typename... Args>
struct eval {
    using type = typename C::template apply<Args...>;
};

template <typename C, typename... Args>
struct curry {
    struct type {
        template <typename... OtherArgs>
        using apply = typename eval<C, Args..., OtherArgs...>::type;
    };
};

两个编译器都编译just fine。原始示例是否有问题,或者这只是两个编译器中的错误?

0 个答案:

没有答案