How to get the first n elements of a std::map

Question

Since there is no .resize() member function in C++ std::map I was wondering, how one can get a std::map with at most n elements.

The obvious solution is to create a loop from 0 to n and use the nth iterator as the first parameter for std::erase().

I was wondering if there is any solution that does not need the loop (at least not in my user code) and is more "the STL way to go".

Answer 1

You can use std::advance( iter, numberofsteps ) for that.

Answer 2

Universal solution for almost any container, such as std::list, std::map, boost::multi_index. You must check the size of your map only.

template<class It>
It myadvance(It it, size_t n) {
   std::advance(it, n);
   return it;
}

template<class Cont>
void resize_container(Cont & cont, size_t n) {
    cont.erase(myadvance(cont.begin(), std::min(n, cont.size())), 
                 cont.end());
}

Answer 3

The correct way for this is to use std::advance. But here is a funny (slow) way allowing to 'use resize on map'. More generally, this kind of trick can be used for other things working on vector but not on map.

map<K,V> m; //your map
vector< pair<K,V> > v(m.begin(), m.end());
v.resize(n);
m = map<K,V>(v.begin(),v.end());

Answer 4

A std::map is not a list. There are no "first n" elements.

BTW: Iterators become invalid if the container is changed.

If you really need a smaller map you could iterate though it and add all elements up to the n-th into a new map.

Answer 5

Why would you want to resize a map?

The elements in a map aren't stored in any order - the first 'n' doesn't really mean anything

edit:
Interestingly std::map does have an order, not sure how useful this concept is.
Are the entries in the same sort order as the keys?
What does that mean? If you have Names keyed by SSN does that mean the names are stored in SSN numeric order?