Re: Multimap: how to get a key list?

tonydee <>
Mon, 1 Mar 2010 19:05:41 -0800 (PST)
On Mar 2, 9:54 am, "Daniel T." <> wrote:

template < typename map_t, typename OutIt>
void keys(const map_t& aMap, OutIt out)
   typename map_t::const_iterator it = aMap.begin();
   while (it != aMap.end()) {
      typename map_t::key_type key = it++->first;
      out++ = key;
      while (it != aMap.end() && key == it->first)

Needs to be "*out++".

I like that you've used an output iterator: set<> was a wasteful way
of handling duplicate keys, even assuming that they're not wanted in
the result (which the OP didn't specify). list<> or vector<> could
reasonably be expected to outperform set<>.

Might want to make key a reference, as copying the key could be
expensive (alternatively, but probably more expensive, keep an
iterator to it). Post-increments involve iterator copies - could also
be slightly more expensive than ++it/++out afterwards, or a
preincrement ala

    while (++it != a.Map.end() and key == it->first)

(The short-circuit logical operator constitutes a sequence point.)

Alternatively, one loop:

    typename map_t::key_type* p_key = NULL;

    for (typename map_t::const_iterator it = aMap.begin();
         it != aMap.end(); ++it)
        if (p_key and *p_key == it->first)
        p_key = &(it->first);
        *out++ = *p_key;

Alternatively, we could shoe-horn an STL algorithm. For example (and
assuming we want each instance of duplicate keys:

    template <class Inserter>
    struct Assign_First : public Inserter
        typedef typename Inserter::container_type container_type;

        Assign_First(container_type& c) : Inserter(c) { }

        Assign_First& operator*() { return *this; }

        template <typename Pair>
        Assign_First& operator=(const Pair& value)
            return *this;


    typedef std::multimap<int, std::string> MM;
    MM mm;

    typedef std::vector<MM::key_type> Keys;
    Keys keys;

    std::copy(mm.begin(), mm.end(),
              Assign_First<std::back_insert_iterator<Keys> >(keys));


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