Question

我正在尝试使用自制的元组实现来学习可变参数模板。有人可以向我解释为什么以下导致编译错误？

namespace my
{
  // Template definition
  template <typename... Ts> struct tuple;

  template <typename T, typename... Ts>
  struct tuple<T, Ts...> : public tuple<Ts...>
  {
    tuple(T t, Ts... ts) : tuple<Ts...>(ts...), mVal(t) {}
    T mVal;
  };

  template <typename T>
  struct tuple<T>
  {
    tuple(T t) : mVal(t) {}
    T mVal;
  };


  // GetType at an index
  template <size_t i, typename... Ts> struct GetType {};

  template <size_t i, typename T, typename... Ts> 
  struct GetType<i, tuple<T,Ts...> > 
  {
    using Type = typename GetType<i-1, tuple<Ts...> >::Type;
  };

  template <typename T, typename... Ts>
  struct GetType<0, tuple<T,Ts...> >
  {
    using Type = T;
  };


  template <size_t i, typename... Ts>
  typename GetType<i,tuple<Ts...> >::Type Get(tuple<Ts...>& t);

  template <size_t i, typename T, typename... Ts>
  typename GetType<i,tuple<T,Ts...> >::Type Get(tuple<T,Ts...>& t)
  {
    return Get<i-1, tuple<Ts...> >(t);
  }

  template <typename T, typename... Ts>
  typename GetType<0,tuple<T,Ts...> >::Type Get(tuple<T,Ts...>& t)
  {
    return (static_cast<tuple<T, Ts...> >(t)).mVal;
  }  
}

int main()
{
  using myTuple = my::tuple<int, std::string, double, char>;

  // The following lines compile fine ....
  my::GetType<0,myTuple>::Type s0 = 437;
  my::GetType<1,myTuple>::Type s1 = std::string("Test string");
  my::GetType<2,myTuple>::Type s2 = 299.3243;
  my::GetType<3,myTuple>::Type s3 = 'Z';
  std::cout << s0 << " - " << s1 << " - " << s2 << " - " << s3 << std::endl;

  myTuple t(437, "This is the actual tuple string", 299.3243, '§');
  // This line does not compile !!! 
  int v = my::Get<0>(t);

  return 0;
}

我的意图是指示的行将使用特化，但查看编译错误很明显，value=0的特化没有被使用。非常感谢您的帮助。

感谢您的时间..

c:\users\praka\workspace\cpp\cpp-tests\recipes\tuple\tuple.h(27): error C2039: 'Type': is not a member of 'my::GetType<18446744073709551614,my::tuple<>>'
c:\users\praka\workspace\cpp\cpp-tests\recipes\tuple\tuple.h(27): note: see declaration of 'my::GetType<18446744073709551614,my::tuple<>>'
c:\users\praka\workspace\cpp\cpp-tests\recipes\tuple\tuple.h(43): note: see reference to class template instantiation 'my::GetType<18446744073709551615,my::tuple<my::tuple<std::string,double,char>>>' being compiled
..\test.cpp(15): note: see reference to function template instantiation 'int my::Get<0,int,std::string,double,char>(my::tuple<int,std::string,double,char> &)' being compiled
c:\users\praka\workspace\cpp\cpp-tests\recipes\tuple\tuple.h(27): error C2061: syntax error: identifier 'Type'
c:\users\praka\workspace\cpp\cpp-tests\recipes\tuple\tuple.h(27): error C2238: unexpected token(s) preceding ';'
c:\users\praka\workspace\cpp\cpp-tests\recipes\tuple\tuple.h(43): error C2672: 'Get': no matching overloaded function found
c:\users\praka\workspace\cpp\cpp-tests\recipes\tuple\tuple.h(43): error C2770: invalid explicit template argument(s) for 'GetType<i,my::tuple<Ts...>>::Type my::Get(my::tuple<Ts...> &)'
c:\users\praka\workspace\cpp\cpp-tests\recipes\tuple\tuple.h(38): note: see declaration of 'my::Get'

Answer 1

关闭主题介绍建议：如果您想使用语言来重新创建标准工具的版本（在本例中为std::tuple），请避免为结构，类，函数指定相同的名称，类型......

这可以避免很多强烈的麻烦。

建议结束。

正如Daniel Langr所观察到的，模板函数没有部分专业化。

因此很难通过函数获得你想要的东西。

但结构/类可以部分专门化，所以我建议编写一个getH助手struct如下

  template <std::size_t I, typename T, typename ... Ts>
  struct getH
   {
     static typename GetType<I, tuple<T, Ts...>>::Type
        func (tuple<T, Ts...> const & t)
      { return getH<I-1U, Ts...>::func(t); }
   };

  template <typename T, typename ... Ts>
  struct getH<0U, T, Ts...>
   {
     static T func (tuple<T, Ts...> const & t)
      { return t.mVal; }
   };

因此可以撰写Get

  template <std::size_t I, typename ... Ts>
  typename GetType<I, tuple<Ts...>>::Type Get(tuple<Ts...> const & t)
   { return getH<I, Ts...>::func(t); };

如果您可以使用C ++ 14，则可以在auto而不是typename GetType<I, tuple<T, Ts...>>::Type中简化返回类型。

我发现在您的代码中，您在不需要时使用了static_cast

您已在Get<0>

中使用过它

  template <typename T, typename... Ts>
  typename GetType<0,tuple<T,Ts...> >::Type Get(tuple<T,Ts...>& t)
  {
    return (static_cast<tuple<T, Ts...> >(t)).mVal;
  }

当mVal类型为T（又名typename GetType<0,tuple<T,Ts...> >::Type）

时

使用Variadic模板的元组编译错误

1 个答案: