递归可变参数模板以打印出参数包的内容 [英] recursive variadic template to print out the contents of a parameter pack

问题描述

如何创建递归可变参数模板以打印出参数包的内容? 我正在尝试使用此方法，但无法编译:

How is it possible to create a recursive variadic template to print out the contents of a paramater pack? I am trying with this, but it fails to compile:

template <typename First, typename ...Args>
std::string type_name () {
    return std::string(typeid(First).name()) + " " + type_name<Args...>();
}
std::string type_name () {
    return "";
}

我应如何结束递归?

推荐答案

您需要使用部分专业化来结束递归，但是由于您不能部分地专业化C ++中的自由函数，因此需要使用静态成员函数.

You need to use partial specialisation to end the recursion, but since you can't partially specialise free functions in C++, you need to create an implementation class with a static member function.

template <typename... Args>
struct Impl;

template <typename First, typename... Args>
struct Impl<First, Args...>
{
  static std::string name()
  {
    return std::string(typeid(First).name()) + " " + Impl<Args...>::name();
  }
};

template <>
struct Impl<>
{
  static std::string name()
  {
    return "";
  }
};

template <typename... Args>
std::string type_name()
{
    return Impl<Args...>::name();
}

int main()
{
  std::cout << type_name<int, bool, char, double>() << std::endl; // "i b c d"
  return 0;
}

Impl的第一个声明只是

That first declaration of Impl is just a workaround for a shortcoming in g++ 4.6 (and below). It won't be necessary once it implements variadic templates correctly.

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