Making a template work with both std::map and std::unordered_map

Question

I have the following template function that returns a copy of the given map with swapped keys and values:

template<typename M>
auto swapKeysAndValues(const M& m) {
    std::map<typename M::mapped_type, typename M::key_type> swapped;
    for (auto& p : m) {
        swapped.emplace(p.second, p.first);
    }
    return swapped;
}

Is there a way of making the above template work for both std::map and std::unordered_map? That is, for std::map<K, V>, it should return std::map<V, K>, and for std::unordered_map<K, V>, it should return std::unordered_map<V, K>.

Answer 1

template<template <typename...> class Map, typename K, typename V>
auto swapKeyValues(const Map<K, V>& map)
{
    Map<V, K> result;
    for (const auto& p : map) result.emplace(p.second, p.first);
    return result;
}

Live example

Answer 2

There are a couple of answers here so I'm not going to cover old ground.

There is however an aspect of this that you ought to consider carefully.

unordered maps are not the same as maps - they have the requirement that a hash function exists for the key (in addition to the equality predicate). As we have seen, it's trivial to write a template function that assumes the defaults, but is that what you want?

If both your K and V have hash functions available, then they are already keys. In that case, isn't what you really wanted a boost::bimap or boost::multi_index_map?