具有多个包的c++模板，通过参数和返回值进行推导

Question

我对 C++ 如何解释以下代码感到有点困惑:

template<class ... ReturnTypes, class ... ParameterTypes>
std::tuple<ReturnTypes...> Method(const ParameterTypes & ... Parameters)
{
    (...)
};

编译以下代码时:

std::tuple<unsigned int> R = Object.Method<unsigned int, unsigned int>(10);

我得到:

error: conversion from 'std::tuple<unsigned int, unsigned int>' to non-scalar type 'std::tuple<unsigned int>' requested
  std::tuple<unsigned int> R = Object.Method<unsigned int, unsigned int>(10);

是否有可能创建一个具有两个参数包的模板方法(在非模板类中)——一个用于返回类型(在元组中)，一个用于参数类型？

Answer 1

可以使用template模板参数来区分这两个参数包:

#include <tuple>

template <class... T> struct typelist { };

template <typename... ReturnTypes,
          template <typename...> class T,
          class ... ParameterTypes>
std::tuple<ReturnTypes...> Method(T<ReturnTypes...>,
                                  const ParameterTypes & ... Parameters)
{
    return std::tuple<ReturnTypes...>();
};

int main()
{
    using ReturnTypes = typelist<int, char>;
    auto t = Method(ReturnTypes{}, 10, "hello", false);

    static_assert(std::is_same<decltype(t), std::tuple<int,char>>{}, "types do not match");
    return 0;
}

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