Question

I want to have a function with interface like this:

template<typename T, typename R> int find_index (const T& list, const R& value);

As I know, there is find() in STL that returns iterator. I need to return index of iterator (even for non-indexed containers such as std::list). I tried this code:

template<typename T, typename R>
int find_index (const T& list, const R& value)
{
    int index = 0;
    for (T::const_iterator it = list.begin(); it != list.end(); it++, index++)
        if ((*it) == value)
            return index;
    return -1;
}

But compiler shows error on it – seems like it is not allowed to get const_iterator from templated typename. Can I go around it?

At the worst case I can pass begin and end iterators to find_index arguments, but it looks not so fine. Would be thankful for elegant solution.

Answer 1

for (typename T::const_iterator it = list.begin(); it != list.end(); ++it, ++index)

should solve your problem.

When using dependent types (types depending on template parameters), the compiler does not know that const_iterator is a type until it instantiates the template with a concrete type, it could also just be a static variable or whatever. Using the typename keyword, you tell him that const_iterator is really a type.

In C++11 you can also circumvent the whole typename issue using the auto keyword:

for (auto it = list.begin(); it != list.end(); ++it, ++index)

---

If you already have the iterator (maybe from some other operation), you can also just compute the distance from the list’s begin to this iterator:

#include <iterator>

int index = std::distance(list.begin(), it);

But since this has linear complexity for a std::list, using your self-made find_index function is a better idea than std::find followed by std::distance, at least performance-wise.