函数基类的不明确重载

Question

我试图overload在http://en.cppreference.com/w/cpp/utility/variant/visit中编写与结构相同的代码,并将其扩展为使用函数.

这是下面转载的代码https://wandbox.org/permlink/5Z2jsEjOewkGoPeX

#include <utility>
#include <type_traits>
#include <cassert>
#include <string>

namespace {
    template <typename Func>
    class OverloadFuncImpl : public Func {
    public:
        template <typename F>
        explicit OverloadFuncImpl(F&& f) : Func{std::forward<F>(f)} {}
        using Func::operator();
    };
    template <typename ReturnType, typename... Args>
    class OverloadFuncImpl<ReturnType (*) (Args...)> {
    public:
        template <typename F>
        explicit OverloadFuncImpl(F&& f) : func{std::forward<F>(f)} {}
        ReturnType operator()(Args... args) {
            return this->func(args...);
        }
    private:
        ReturnType (*func) (Args...);
    };

    template <typename... Funcs>
    class Overload;
    template <typename Func, typename... Funcs>
    class Overload<Func, Funcs...>
            : public OverloadFuncImpl<Func>,
              public Overload<Funcs...> {
    public:
        template <typename F, typename... Fs>
        explicit Overload(F&& f, Fs&&... fs)
            : OverloadFuncImpl<Func>{std::forward<F>(f)},
            Overload<Funcs...>{std::forward<Fs>(fs)...} {}
        using OverloadFuncImpl<Func>::operator();
        using Overload<Funcs...>::operator();
    };
    template <typename Func>
    class Overload<Func> : public OverloadFuncImpl<Func> {
    public:
        template <typename F>
        explicit Overload(F&& f) : OverloadFuncImpl<Func>{std::forward<F>(f)} {}
        using OverloadFuncImpl<Func>::operator();
    };
}

template <typename... Funcs>
auto make_overload(Funcs&&... funcs) {
    return Overload<std::decay_t<Funcs>...>{std::forward<Funcs>(funcs)...};
}

char foo(char ch) {
    return ch;
}

int main() {
    auto overloaded = make_overload(
        [&](int integer) { return integer; },
        [&](std::string str) { return str; },
        [&](double d) { return d; },
        foo);

    assert(overloaded("something") == "something");
    assert(overloaded(1.1) == 1.1);

    return 0;
}

这是我得到的错误

In file included from /opt/wandbox/gcc-7.2.0/include/c++/7.2.0/cassert:44:0,
                 from prog.cc:3:
prog.cc: In function 'int main()':
prog.cc:66:26: warning: ISO C++ says that these are ambiguous, even though the worst conversion for the first is better than the worst conversion for the second:
     assert(overloaded(1.1) == 1.1);
                          ^
prog.cc:62:21: note: candidate 1: main()::<lambda(double)>
         [&](double d) { return d; },
                     ^
prog.cc:19:20: note: candidate 2: ReturnType {anonymous}::OverloadFuncImpl<ReturnType (*)(Args ...)>::operator()(Args ...) [with ReturnType = char; Args = {char}]
         ReturnType operator()(Args... args) {
                    ^~~~~~~~

编译器和标准解释存在一些问题,使得必须operator()逐个导入.但不知何故operator(),功能专业化OverloadFuncImpl似乎没有通过using正确导入.

请注意,当我不使用OverloadFuncImpl或排除函数部分特化时,上面的代码工作得很好OverloadFuncImpl.

我已经得到了这个代码与一个解决方法,但我只是想知道为什么上面的代码不起作用.我似乎无法弄明白......为什么我导入了operator()所有的基类.仍存在模糊的过载问题？

我试图在较小的上下文中重现错误,但无法...

Answer 1

    ReturnType operator()(Args... args) const {
//                                      ^^^^^
        return this->func(args...);
    }

有效地解决了过载问题中的相关候选人

 char operator()(char);
 double operator()(double) const;

调用const带有类型参数的非对象double.

第一个赢得隐式对象参数; 第二个赢得实际的函数参数.随之而来的是歧义.