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我一直在尝试任何我能想到的让_CallWithRightmostArgsInner函数正确失败的方法，以便 SFINAE 可以正常工作，通过这次尝试，VS2013 给了我错误： error C2039: 'type' : is not a member of 'std::enable_if<false,void>'

有任何想法吗？有没有更好的替代方案？这里的想法是，我想对 Function 进行函数调用，前提是 Function 采用 NumArgs 表示的数字或参数。最后两个可变参数应转发给函数并返回结果。

template <typename Function, int NumArgs>
class SplitParameters {
public:
    typedef typename function_traits<Function>::result_type result_type;

    template <typename ... RightArgs>
    static result_type CallWithRightmostArgs(const Function& call, RightArgs && ... rightArgs) {
        static_assert(sizeof...(RightArgs) >= NumArgs, "Unable to make function call with fewer than minimum arguments.");
        return _CallWithRightmostArgs(call, std::forward<RightArgs>(rightArgs)...);
    }

private:
    template <typename ... RightArgs>
    static result_type _CallWithRightmostArgs(const Function& call, RightArgs && ... rightArgs) {
        return _CallWithRightmostArgsInner(call, std::forward<RightArgs>(rightArgs)...);
    }

    // note the '==' vs '!=' in these two functions.  I would assume that only one could exist
    template <typename LeftArg, typename ... RightArgs, typename std::enable_if<sizeof...(RightArgs) != NumArgs>::type* = 0>
    static result_type _CallWithRightmostArgsInner(const Function& call, LeftArg, RightArgs && ... rightArgs) {
        return _CallWithRightmostArgs(call, std::forward<RightArgs>(rightArgs)...);
    }

    template <typename LeftArg, typename ... RightArgs, typename std::enable_if<sizeof...(RightArgs) == NumArgs>::type* = 0>
    static result_type _CallWithRightmostArgsInner(const Function& call, LeftArg, RightArgs && ... rightArgs) {
        return call(std::forward<RightArgs>(rightArgs)...);
    }
};
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1 回答 1

3

我通过将您的代码更改为 g++-4.8

    #include <iostream>

    template <class T>
    struct function_traits
    {
        typedef void result_type;
    };

    template <typename Function, int NumArgs>
    class SplitParameters {
    public:
        typedef typename function_traits<Function>::result_type result_type;

        template <typename ... RightArgs>
        static result_type CallWithRightmostArgs(const Function& call, RightArgs && ... rightArgs) {
            static_assert(sizeof...(RightArgs) >= NumArgs, 
                          "Unable to make function call with fewer than minimum arguments.");
            return _CallWithRightmostArgs(call, std::forward<RightArgs>(rightArgs)...);
        }

    private:
        template <typename ... RightArgs>
        static result_type _CallWithRightmostArgs(const Function& call, RightArgs && ... rightArgs) {
            return _CallWithRightmostArgsInner(call, std::forward<RightArgs>(rightArgs)...);
        }

        // note the '==' vs '!=' in these two functions.  I would assume that only one could exist
        template <typename LeftArg, typename ... RightArgs, class = typename std::enable_if<sizeof...(RightArgs) != NumArgs -1 >::type>
        static result_type _CallWithRightmostArgsInner(const Function& call, LeftArg, RightArgs && ... rightArgs) {
            return _CallWithRightmostArgsInner(call, std::forward<RightArgs>(rightArgs)...);
        }

        template <typename ... RightArgs, class = typename std::enable_if<sizeof...(RightArgs) == NumArgs>::type>
        static result_type _CallWithRightmostArgsInner(const Function& call, RightArgs && ... rightArgs) {
            return call(std::forward<RightArgs>(rightArgs)...);
        }
    };

    void f(int i, int j)
    {
        std::cout << i << ' ' << j << std::endl;
    }

    int main()
    {
        SplitParameters<decltype(f), 2>::CallWithRightmostArgs(f, 1, 2, 3, 4);
    }

编译器不喜欢您调用_CallWithRightmostArgsfrom _CallWithRightmostArgsInner，我假设您实际上是在尝试调用该Inner函数。
g++ 也不喜欢在模板参数列表中转换0为void*，所以我将其更改为class = enable_if<...>::type。

我没有详细研究它失败的原因，希望这对你来说已经足够好了。

编辑：关于typename enable_if<...>::type* = 0被拒绝，我记得有一个类似的问题std::array：

    template <class T, int size>
    void f(const std::array<T,size>&){}

这个小片段自己编译就好了，但是当你这样做时：

    std::array<int,4> a;
    f(a);

    g++ gives:
    test3.cpp: In function ‘int main()’:
    test3.cpp:9:8: error: no matching function for call to ‘f(std::array<int, 4ul>&)’
         f(a);
            ^
    test3.cpp:9:8: note: candidate is:
    test3.cpp:4:6: note: template<class T, int size> void f(const std::array<T, size>&)
     void f(const std::array<T,size>&){}
          ^
    test3.cpp:4:6: note:   template argument deduction/substitution failed:
    test3.cpp:9:8: note:   mismatched types ‘int’ and ‘#‘integer_cst’ not supported by dump_type#<type error>’
         f(a);
            ^
    test3.cpp:9:8: note:   ‘std::array<int, 4ul>’ is not derived from ‘const std::array<T, size>’

事实证明，问题是我将模板声明int为size参数，但编译器得到的是 astd::size_t与 an 不同int，即使您可以轻松地在它们之间进行转换。
在上面的示例中，我什至无法替换= 0为，= NULL因为这只是一个0L文字，我必须= (void*)0让编译器接受它（因为默认类型enable_if<true>::typeis void）。

于 2013-09-11T17:45:50.400 回答