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copied!<p>The reason is that your <code>K</code> and <code>V</code> type parameters are in a non-deducible context, so your function template is never even instantiated during overload resolution. </p> <p>Look at it again:</p> <pre><code>template <typename K, typename V> map_keys_iterator<K,V> map_keys(const typename std::map<K,V>::iterator &i) </code></pre> <p>For this to work, the C++ compiler would somehow have to walk from a specific <code>iterator</code> class to its "parent container" type - <code>map</code> in this case - to get its <code>K</code> and <code>V</code>. In general, this is impossible - after all, a particular <code>iterator</code> might be a typedef for some other class, and the actual type of argument in the call is that other class; there's no way the compiler can "retrace" it. So, per C++ standard, it doesn't even try in this case - and, more generally, in any case where you have <code>typename SomeType<T>::OtherType</code>, and <code>T</code> is a type parameter.</p> <p>What you can do is make the entire parameter type a template type parameter. This requires some trickery to derive <code>K</code> and <code>V</code>, though.</p> <pre><code>template <typename Iterator> map_keys_iterator< typename std::iterator_traits<Iterator>::value_type::first_type, typename std::iterator_traits<Iterator>::value_type::second_type > map_keys(Iterator i) </code></pre> <p>Unfortunately, you'll have to repeat those two in the body of the function as well, when invoking the constructor of your type.</p> <p>As a side note, iterators are generally passed by value (they're meant to be lightweight).</p>

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