在泛型类中获取泛型容器的迭代器:如何

getting an iterator to a generic container inside a generic class: HOW TO

本文关键字：迭代器 如何 泛型类 获取 泛型      更新时间：2023-10-16

这来自一个更大的上下文，但是我已经剥离了许多代码来简化问题。如果你觉得我遗漏了什么，请告诉我。

假设你有一个模板类定义为:

#include <map>
#include <vector>
#include <string>
#include <fstream>
#include <iostream>
#include <iterator>
#include <algorithm>
template <class key, class container, class container_type>
class file_to_map
{
public:
  file_to_map()
  {
    m_file = "";
  }
  file_to_map(std::string file)
  {
    m_file = file;
  }
  ~file_to_map()
  {
  }
  std::map<key, container>& get_map()
  {
    return m_map;
  }
  void set_file(const std::string file)
  { 
    m_file = file;
  }
  void insert_into_map(key insert, container_type value)
  {
    m_map[insert].insert(value);
  }
  friend std::ostream& operator<< (std::ostream &out, file_to_map<key, container, container_type> &obj)
  {
    typedef typename std::map<key, container>::const_iterator mapItr;
    mapItr mbi = obj.m_map.begin();
    mapItr emi = obj.m_map.end();
    while (mbi != emi) {
      out << " -- " << mbi->first << " -- " << std::endl;
      container::iterator cbi;
      ++mbi;
    }
    return out;
  }
  friend std::istream& operator>> (std::ifstream &in, file_to_map<key, container, container_type> &obj)
  {
    if (in.is_open())
      in.close();
    if (obj.m_file == "")
      return in;
    in.open(obj.m_file.c_str(), std::ios::in);
    if (in.fail() || in.bad()) {
      in.close();
      return in;
    }
    std::vector<key> tmp;
    typedef std::istream_iterator<key> string_input;
    copy(string_input(in), string_input(), back_inserter(tmp));
    typename std::vector<key>::iterator bvi = tmp.begin();
    typename std::vector<key>::iterator evi = tmp.end();
    while (bvi != evi) {
      obj.m_map[*(bvi)] = container();
      ++bvi;
    }
    in.close();
    return in;
  }  
  private:
    std::map<key, container> m_map;
    std::string m_file;
};

和友元方法" operator<< "

你想打印地图的内容和通用的container怎么做呢?如何获得一个合适的迭代器来遍历泛型container .

我正在尝试:

container::iterator cbi;

代码更新

: 

friend std::ostream& operator<< (std::ostream &out, file_to_map<key, container, container_type> &obj)
  {
    typedef typename std::map<key, container>::const_iterator mapItr;
    mapItr mbi = obj.m_map.begin();
    mapItr emi = obj.m_map.end();
    while (mbi != emi) {
      out << " -- " << mbi->first << " -- " << std::endl;
      typename container::const_iterator cbi = mbi->second.begin();
      typename container::const_iterator ebi = mbi->second.end();
      std::copy(cbi, mbi, std::ostream_iterator<container_type>(out, "tn"));
      ++mbi;
    }
    return out;
  }

我得到以下编译错误:

g++ -o file_to_map -Wall ./file_to_map.h ./main.cpp ./code_finder.cpp 
        -L/usr/local/boost_1_48_0/stage/lib -lboost_filesystem -lboost_system -I /usr/local/boost_1_48_0/
In file included from ./code_finder.h:4,
                 from ./code_finder.cpp:1:
./file_to_map.h: In function ‘std::ostream& operator<<(std::ostream&, file_to_map<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::set<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::less<std::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::allocator<std::basic_string<char, std::char_traits<char>, std::allocator<char> > > >, std::basic_string<char, std::char_traits<char>, std::allocator<char> > >&)’:
./code_finder.cpp:36:   instantiated from here
./file_to_map.h:62: error: no matching function for call to ‘copy(std::_Rb_tree_const_iterator<std::basic_string<char, std::char_traits<char>, std::allocator<char> > >&, operator<<(std::ostream&, file_to_map<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::set<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::less<std::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::allocator<std::basic_string<char, std::char_traits<char>, std::allocator<char> > > >, std::basic_string<char, std::char_traits<char>, std::allocator<char> > >&)::mapItr&, std::ostream_iterator<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, char, std::char_traits<char> >)’

我错过了什么?

使用typename:

typename container::iterator cbi;

因为iterator是一个依赖名称，因为它依赖于类模板的类型参数container。有趣的是，如果你在其他地方使用了typename，比如operator<<:

typedef typename std::map<key, container>::const_iterator mapItr;

因为你正在使用const_iterator的映射，这意味着容器对象，你会得到使用map的常量迭代器将是const的对象，这反过来意味着你需要使用const_iterator的容器，因为它们是映射的值。所以我认为你需要使用这个:

typename container::const_iterator cbi; //use this instead!

回复你的编辑:

我看到这里有个错别字:

typename container::const_iterator cbi = mbi->second.begin();
typename container::const_iterator ebi = mbi->second.end();
std::copy(cbi, mbi, std::ostream_iterator<container_type>(out, "tn"));
            // ^^^ typo

std::copy的第二个参数应该是ebi，而不是mbi。这就是为什么我通常将这些变量命名为begin和end，而不是cbi和ebi。

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