3

这是我的代码，作为背景：

template<typename... Dummy>
struct tuple;

template <typename T, typename... TList>
struct tuple<T,TList...>
    :public tuple<TList...>
{
    typedef T value_type;

    tuple(){}

    template<typename A,typename... AList>
    tuple(const A& a,const AList&... args)
        :tuple<TList...>(args...),element(a)
    {
    }

    template<typename A,typename... AList>
    tuple(A&& a,AList&&... args)
        :tuple<TList...>(args...),element(a)
    {
    }

    tuple(const tuple<T,TList...>& a)
        :tuple<TList...>(a),element(a.element)
    {
    }

    tuple(tuple<T,TList...>&& a)
        :tuple<TList...>(a),element(a.element)
    {
    }

    static const index_t size=1+tuple<TList...>::size;
    static const index_t offset=sizeof(tuple<TList...>);

    tuple_unit<T> element;
};

template<typename T>
struct tuple<T>
{
    typedef T element_type;

    tuple(){}

    template<typename A>
    tuple(const A& a)
        :element(a)
    {
    }

    template<typename A>
    tuple(A&& a)
        :element(a)
    {       
    }

    tuple(const tuple<T>& a)
        :element(a.element)
    {
    }

    tuple(tuple<T>&& a)
        :element(a.element)
    {
    }

    static const index_t size=1;
    static const index_t offset=0;

    tuple_unit<T> element;
};

由于在 C++11 中我们有移动语义，我尝试在我的项目中添加移动构造函数。但结果并不如我所料。

当我写

tuple<int,float,double> a(3,5,7); //Success. Use 3,5,7 to init each element in tuple.
tuple<int,float,double> b(a); //Call template function,not copy ctor,and it failed.

我读了这个，它让我认为b(a)会调用template<typename A,typename... AList> tuple(A&& a,AList&&... args)，并且A& &&会被替换A&，但我已经有一个构造函数tuple(const tuple<T>& a)。为什么编译器会认为模板构造函数比复制构造函数好？

我应该怎么做才能解决问题？

4

1 回答 1

1

我真的不认为构造函数本身也需要模板，见下文。此外，在初始化元组成员时，请考虑在移动构造函数中使用“std::move”。

template<typename... Dummy>
struct tuple;

template <typename T, typename... TList>
struct tuple<T,TList...>
  :public tuple<TList...>
{
  typedef T value_type;

  tuple(){}

  tuple(const T& a,const TList&... args)
    :tuple<TList...>(args...),element(a)
  {
  }

  tuple(T&& a,TList&&... args)
    :tuple<TList...>(args...),element(a)
  {
  }

  tuple(const tuple<T,TList...>& a)
    :tuple<TList...>(a),element(a.element)
  {
  }

  tuple(tuple<T,TList...>&& a)
    :tuple<TList...>(a),element(a.element)
  {
  }

  static const int size=1+tuple<TList...>::size;
  static const int offset=sizeof(tuple<TList...>);

  T element;
};

template<typename T>
struct tuple<T>
{
  typedef T element_type;

  tuple(){}

  tuple(const T& a)
    :element(a)
  {
  }

  tuple(T&& a)
    :element(a)
  {       
  }

  tuple(const tuple<T>& a)
    :element(a.element)
  {
  }

  tuple(tuple<T>&& a)
    :element(a.element)
  {
  }

  static const int size=1;
  static const int offset=0;

  T element;
};

int
main ()
{
  tuple<int,float,double> a(3,5,7); //Success. Use 3,5,7 to init each element in tuple.

  tuple<int,float,double> b(a); //Call template function,not copy ctor,and it failed.
}
于 2013-09-08T16:19:40.573 回答