Question

我想知道我在哪里可以做这样的事情（c ++ 11）：

//class FOO is a static class which contains static members and lots of typedefs for 
container types such as vector, map, BOOST Graph (adjacency list), property_maps etc


 current implementation:

// file: traits.hpp
  class FOO {
    typedef std::vector<int> container_t;
    typedef std::map<int, int>  map_t;
    typedef boost::adjacency_list < > graph_t;
    typedef boost::property_map <  >  p_map_t  // syntax is not correct though(just for understanding purpose)

    // lots of other typedefs ...
    static int var1;
    static int var2;    // lots of other static variables 
  }; 
    int FOO::var1 = 10;
    int FOO::var2 = 20; //initialize static variables 


 // file BAR.hpp     
 template <class FOO>
 class BAR {
   public: 
     typedef typename FOO::map_t        map_t;
     typedef typename FOO::container_t  conatiner_t;  // and so on
     typedef typename FOO::graph_t      graph_t 


     BAR () { ... constructor ... }

   private:
     map_t        my_map;
     container_t  my_container;
     graph_t      my_Graph;
  };

Problem in above: lots of typedef typename need to be used always. Note that I 
could directly use this in private: e.g. 
  typename FOO::map_t  my_map;
  typename FOO::graph_t  my_graph;

  but am avoiding it as there are lots of typenames and stuff which i will have 
  to manage and the code


//What I want to do :-->


// file: traits.hpp

// create a namespace "traits"
namespace traits {
  class FOO {
    typedef std::vector<int> container_t;
    typedef std::map<int, int>  map_t;
    typedef boost::adjacency_list < > graph_t;
    // lots of other typedefs
    static int var1;
    static int var2;    // lots of other static variables 
  }; 
    int FOO::var1 = 10;
    int FOO::var2 = 20; //initialize static variables 
 }; // end namespace


 // file BAR.hpp 

 using namespace traits;

 class BAR {
   public: 
       template <..>
       void bar ()  // functions
   private:
     FOO::map_t  my_map;
     FOO::container_t  my_container;
     FOO::graph_t      my_Graph;
  }

  OR  would be like this :

   private :
     traits::FOO::map_t my_map;
     traits::graph_t    my_graph;
     ... 

   Cant I directly use like this as FOO will be the only class in namespace traits?
     traits::map_t  my_map;
     traits::graph_t my_graph;  //and so on...

我的主要动机是将所有typedef和静态变量保存在类（FOO）中并将其包装在命名空间（traits）中，这样我就不必将类FOO作为模板参数显式传递给类BAR，并且然后再次使用typedef，它不会使代码看起来很好，因为我有很多其他类使用FOO容器。同时我需要将容器typedef和static变量封装在一个类中。

有人可以帮帮我吗？谢谢！

Answer 1

只需为特征创建自定义命名空间：

namespace traits
{
    typedef std::vector<int> container_t;
    typedef std::map<int, int>  map_t;
    typedef boost::adjacency_list < > graph_t;
    typedef boost::property_map <  >  p_map_t  // syntax is not correct though(just for understanding purpose)
};


class BAR {
public: 
       template <..>
       void bar ()  // functions
private:
     traits::map_t  my_map;
     traits::container_t  my_container;
     traits::graph_t      my_Graph;

.... 
};

创建命名空间类以在另一个类中定义容器

1 个答案: