Question

Consider a std::map<K,V>. I want to re-order the map by value profiting by an appropriate container std::C<V*> or std::C<V&>, in a way that no copies of values are done to store the elements in C. Furthermore, elements in C must be sorted according to the result of int f(V&) applied to each element. Despite my efforts I could not find an appropriate C and an enough efficient way to build it. Do you have any solution? A small example would be much appreciated.

Answer 1

Seems simple enough.

std::map<K,V> src;
int f(V&) {return 0;}

V* get_second(std::pair<const K,V> &r) {return &(r.second);} //transformation
bool pred(V* l, V* r) { return f(*l)<f(*r); }   //sorting predicate

std::vector<V*> dest(src.size());  //make destination big enough
std::transform(src.begin(), src.end(), dest.begin(), get_second); //transformcopy
std::sort(dest.begin(), dest.end(), pred); //sort

Unless you meant C is supposed to be another map:

std::pair<K,V*> shallow_pair(std::pair<const K,V> &r) 
{return std::pair<K,V*>(r.first, &(r.second));}

std::map<K, V*> dest2;
std::transform(src.begin(), src.end(), 
               std::inserter(dest2,dest2.end()), shallow_pair);

http://ideone.com/bBoXq

This requires the previous map to remain in scope longer than dest, and have no pairs removed until dest is destructed. Otherwise src will need to have been holding smart pointers of some sort.