What does "typename iterator_traits::difference_type" mean?

Question

This is the example-implementation of the count algorithm from https://devdocs.io/cpp/algorithm/count_if:

template<class InputIt, class T>
typename iterator_traits<InputIt>::difference_type
    count(InputIt first, InputIt last, const T& value)
{
    typename iterator_traits<InputIt>::difference_type ret = 0;
    for (; first != last; ++first) {
        if (*first == value) {
            ret++;
        }
    }
    return ret;
}

My question is, what is the significance of typename iterator_traits<InputIt>::difference_type?
If I were implementing this, I would have simply used unsigned int to keep track of the count.

score 4 · Accepted Answer · answered Mar 04 '20 at 05:38

4

Well, you cannot know the best sufficiently-large type the difference between two iterators without knowing anything about the iterators.
As an example, what if the iterator iterates the bytes in a file:
The filesize is 64bit, but we are in a 32bit process. std::size_t won't work, likely unsigned won't either.

Thus, ask std::iterator_traits to generically provide a suitable type.

Now, we have to use typename there to assure the compiler that the dependent qualified name std::iterator_traits<InputIt>::difference_type will be a type. That is important to clarify for 2-phase lookup.

answered Mar 04 '20 at 05:38

Deduplicator

44,692
7
66
118

So you're saying that unsigned int might not be large enough to hold the count? – izaguirrejoe Mar 04 '20 at 15:47
1

It might not be enough. It might also be more than needed. – Deduplicator Mar 04 '20 at 16:01

score 0 · Answer 2 · answered Mar 04 '20 at 05:41

0

typename tells the compiler that iterator_traits::difference_type is of type class.

Consider an easier to understand example:

class ...
{
    typename T::Something *p;
}

Without the typename, the compiler may try to create a static member which will contain the result of T::Something multiplied by p;

while when you state typename, the compiler is certain to know that p is a pointer of type T::Something

Hope this helps.

answered Mar 04 '20 at 05:41

Lior

284
1
6

I think the question is more about "`difference_type`" than "`typename`". (Note the "If I were implementing this" part of the question.) – JaMiT Mar 04 '20 at 06:08

What does "typename iterator_traits::difference_type" mean?

2 Answers2