Get index in vector from reverse iterator

I would use:

#include <algorithm>
#include <iostream>
#include <vector>

int main()
{
    auto v = std::vector<int> { 1, 2, 3 };
    auto rit = std::find(v.rbegin(), v.rend(), 3);
    if (rit != v.rend()) {
        auto idx = std::distance(begin(v), rit.base()) - 1;
        std::cout << idx;
    } else
        std::cout << "not found!";
}

Live Example.

The reason for the -1 in the distance computation is because of the conversion between reverse and regular iterators in the .base() member:

24.5.1 Reverse iterators [reverse.iterators]

1 Class template reverse_iterator is an iterator adaptor that iterates from the end of the sequence defined by its underlying iterator to the beginning of that sequence. The fundamental relation between a reverse iterator and its corresponding iterator i is established by the identity: &*(reverse_iterator(i)) == &*(i - 1).

Note: you could also use the above code without the check for v.rend(), and use the convention that idx == -1 is equivalent to an element that is not found. However, that loses the ability to do v[idx], so eventually you would need a check against that as well.

You could use:

container.size() - 1 - (iterator - container.rbegin())

or

container.size() - 1 - std::distance(container.rbegin(), iterator)

More info about reverse iterators. How To Use Reverse Iterators Without Getting Confused. To convert reverse iterators into forward iterators and much more.

I would change TemplateRex's answer to use only reverse iterators, thus avoiding any headache with reverse-forward conversions.

int main()
{
    auto v = std::vector<int> { 1, 2, 3 };
    auto rit = std::find(v.rbegin(), v.rend(), 3);
    if (rit != v.rend()) {
        auto idx = std::distance(rit, v.rend()) - 1;
        std::cout << idx;
    } else
        std::cout << "not found!";
}

The -1 is still needed because the first element of the vector (index 0) is actually at v.rend() - 1.