Re: Copying from maps to lists or vectors

From:
James Kanze <james.kanze@gmail.com>
Newsgroups:
comp.lang.c++
Date:
18 Apr 2007 02:15:59 -0700
Message-ID:
<1176887759.264657.52570@y80g2000hsf.googlegroups.com>
On Apr 17, 8:55 pm, pallav <pallavgu...@gmail.com> wrote:

I have a map like this:

typedef boost::shared_ptr<Node> NodePtr;
typedef std::vector<NodePtr> NodeVecPtr;
typedef std::map<std::string, NodePtr> NodeMap;
typedef std::map<std:string, NodePtr>::iterator NodeMapItr;

NodeMap nmap;

In my classes, I often find myself copying the second arguments of
maps to lists/vectors like this.

NodeVecPtr nodes;
for (NodeMapItr itr = nmap.begin(); itr != nmap.end(); ++itr)
nodes.push_back(itr->second);

I may also have node be a list of NodePtrs as opposed to vectors. I'm
wondering if there is some standard STL routines, I can use to try to
reduce this kind of code being repeated. I now std::copy() exists but
it takes iterators. But since I'm dealing with maps, I'm not sure how
to pass the second value as an iterator. There is also
std::transform() but that also works with iterators.

I've been searching the web, and found stuff like boost::bind and
boost::iterator_adaptors that might work but I'm trying to read up the
documentation to figure out the syntax.

If somebody can help me in showing how I could accomplish this using
STL/boost features, this will be helpful. Thanks for you time.


Wouldn't something like the following work:

    template< typename Map >
    struct ValueFromMap
        : public std::unary_function< typename Map::value_type,
                                      typename Map::mapped_type >
    {
        typename Map::mapped_type operator()(
            typename Map::value_type const&
                                elem ) const
        {
            return elem.second ;
        }
    } ;

    typedef boost::transform_iterator< ValueFromMap< NodeMap >,
                                       NodeMap::const_iterator >
ValueIter ;

(I tried it with a std::map< std::string, int >, and it seemed
to work.)

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