In a boost multi_index container, is a "default index" defined?

Question

Here is a container of ints with a sequence index and a hashed index:

#include <iostream>
#include <boost/multi_index_container.hpp>
#include <boost/multi_index/hashed_index.hpp>
#include <boost/multi_index/identity.hpp>
#include <boost/multi_index/sequenced_index.hpp>

int main()
{
    namespace bmi = boost::multi_index;
    boost::multi_index_container<
        int,
        bmi::indexed_by<
            bmi::sequenced<>,
            bmi::hashed_unique<bmi::identity<int>>
        >
    > c;
    for (int i=0; i<100; ++i) c.push_back(i);
    for (int j : c) std::cout << " " << j;
    std::cout << std::endl;
    return 0;
}

Note I did not use get in the second loop. Is the behavior defined in this case? (E.g., "This is the same as using .get<0>()".)

Answer 1

Yes, index #0 is the default in the sense explained here:

The functionality of index #0 can be accessed directly from a multi_index_container object without using get<0>(): for instance, es.begin() is equivalent to es.get<0>().begin().