std::vector move reverse iterator at end

Question

I have this situation:

for(auto it = vec.rbegin(); it != vec.rend(); ++it)
{
    if(condition(it))
    {
        //Move it at end;
        break;
    }
}

What is the most efficient/elegant way to move *it at the end of vec?

EDIT: *it, not it

EDIT1: Now I use:

auto val = *it; 
vec.erase((++it).base());
vec.push_back(val);

But I don't think is very efficient...

Why? You are breaking out of the loop, and `it` will go out of scope anyway. — Femaref, Feb 23 '13 at 19:40
So what do you mean: _swap_ `*it` with `vec.last()`, or "remove it" so that every element after it goes one place further to the front and then append the element, or what? — leftaroundabout, Feb 23 '13 at 19:48
@leftaroundabout I don't want swap, I want the to have the order preserved for the other elements, but to have my element at the end — Mircea Ispas, Feb 23 '13 at 20:06

Bernhard Barker · Accepted Answer · 2013-02-23T20:21:39.017

3

To move it to the end:

for(auto it = vec.rbegin(); it != vec.rend(); ++it)
{
    if(condition(it))
    {
        auto val = *it;
        vec.erase((++it).base());
        vec.push_back(val);
        break;
    }
}

To swap with the last element:

for(auto it = vec.rbegin(); it != vec.rend(); ++it)
{
    if(condition(it))
    {
        auto it2 = --vec.end();
        auto val = *it;
        *it = *it2;
        *it2 = val;
        break;
    }
}

edited Feb 23 '13 at 20:21

answered Feb 23 '13 at 19:48

Bernhard Barker

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And if `it` becomes invalidated by the call to `push_back`? – Benjamin Lindley Feb 23 '13 at 19:50
Not really, because the same conditions that would cause `it` to become invalidated would cause `val` to become invalidated. – Benjamin Lindley Feb 23 '13 at 19:52
Yes, it will compile. It will give you a reference to the thing that `it` points to. – Benjamin Lindley Feb 23 '13 at 19:53
@Dukeling it is reverse iterator and erase works on iterators. The code is something like: auto val = *it; vec.erase((++it).base()); vec.push_back(val); but I don't think is very efficient – Mircea Ispas Feb 23 '13 at 19:54
@Felics Why `vec.erase((++it).base());` not `vec.erase(it.base());`? I doubt it can be more efficient. To remove it from it's current position will take as long as `erase` takes and `push_back` takes constant time. – Bernhard Barker Feb 23 '13 at 20:01
@Dukeling Here is the answer: http://stackoverflow.com/questions/1830158/how-to-call-erase-with-a-reverse-iterator Please correct the erase(it) to be able to accept your answer – Mircea Ispas Feb 23 '13 at 20:12
@Felics Thanks, a split-second look at the picture in the link and it all made sense. Corrected. – Bernhard Barker Feb 23 '13 at 20:22
@BenjaminLindley I see you deleted your answer. I think it was still useful for reference sake. – Bernhard Barker Feb 23 '13 at 20:25

jrok · Answer 2 · 2013-02-23T20:35:03.117

If prefer standard algorithms over manual loops. So I'd use std::find_if.

If you need to preserve the order of the elements, moving to the end can be done using std::rotate:

auto rit = std::find_if(vec.rbegin(), vec.rend(), [](int i){ return i == 6; });
if (rit != vec.rend()) {
    auto it = rit.base();
    std::rotate(it-1, it, vec.end());
}

If you don't need to keep the order, you can use std::iter_swap:

auto it = std::find_if(vec.rbegin(), vec.rend(), [](int i){ return i == 6; });
if (it != vec.rend())
    std::iter_swap(it, vec.rbegin());

Demo

std::vector move reverse iterator at end

2 Answers2