Question

I have a Map::const_iterator object in C++ and i am trying to abstract the entire map object from it without explicitly creating a map with Map::const_iterator->first and Map::const_iterator->second.

this doesn’t seem to work : (*Map::const_iterator).as_map()
please advice.

Answer 1

I’m afraid I understood the question totally wrong. How about using it->second?

 typedef .... mymap_t;
 typedef std::map<std::string, mymap_t> supermap_t;

 supermap_t supermap;


 for (supermap_t::const_iterator it = supermap.begin();
          it != supermap.end();
          ++it)
 {
     const mymap& the_sub_map = it->second;
     call_some_other_function(the_sub_map); // the whole submap
 }

_{From the question it is hard to make out what you actually need, but I’m going to guess that you are looking for a way to treat an iterator range as a container, without having to copy the container.}

Have a look at Boost Range for that.

 typedef map<int, string> mymap_t;

 mymap_t mymap;
 mymap[1] = "One";
 mymap[2] = "Two";
 mymap[3] = "Three";

 mymap_t::const_iterator first(my_map.begin()), 
                         last(my_map.end());

 boost::sub_range<mymap_t> subrange(first++, last); // contains two items

You could treat subrange as a ‘container’ class with boost range algorithms.

Does this come close to what you wanted to ask?

Update

It seems you are after this:

 typedef map<int, string> mymap_t;

 void somefunction(mymap_t::const_iterator first, mymap_t::const_iterator last)
 {
     mymap_t subclone(first, last);

     // use subclone -- it is a mymap_t now :)
 } 

 mymap_t mymap;
 mymap[1] = "One";
 mymap[2] = "Two";
 mymap[3] = "Three";

 somefunction(mymap.begin(), mymap.end());