Iterator to last element in std::list

Question

#include <list>
using std::list;

int main()
{
    list <int> n;
    n.push_back(1);
    n.push_back(2);
    n.push_back(3);

    list <int>::iterator iter = n.begin();
    std::advance(iter, n.size() - 1); //iter is set to last element
}

is there any other way to have an iter to the last element in list?

for some reason, I don't want to have a reverse_iterator. – cpx Apr 20 '10 at 19:58 — cpx, Apr 20 '10 at 19:58

score 114 · Accepted Answer · answered Apr 20 '10 at 19:55

114

Yes, you can go one back from the end. (Assuming that you know that the list isn't empty.)

std::list<int>::iterator i = n.end();
--i;

answered Apr 20 '10 at 19:55

CB Bailey

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In places this can be shortened as in `newlist.splice(--newlist.end(),oldlist);` – dmckee --- ex-moderator kitten Dec 08 '11 at 20:02

Remy Lebeau · Answer 2 · 2014-10-11T17:24:56.203

84

Either of the following will return a std::list<int>::iterator to the last item in the list:

std::list<int>::iterator iter = n.end();
--iter;

std::list<int>::iterator iter = n.end();
std::advance(iter, -1);

// C++11
std::list<int>::iterator iter = std::next(n.end(), -1);

// C++11
std::list<int>::iterator iter = std::prev(n.end());

The following will return a std::list<int>::reverse_iterator to the last item in the list:

std::list<int>::reverse_iterator iter = std::list::rbegin();

edited Oct 11 '14 at 17:24

answered Apr 20 '10 at 19:59

Remy Lebeau

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Will `std::list::iterator iter = --n.end();` work as well or the decrement will be done only after the assignment? – mbrandalero May 02 '18 at 11:20
1

@mbrandalero it should work fine. The result of the decrement will be assigned to the variable – Remy Lebeau May 02 '18 at 14:45
@RemyLebeau That might not always work (i.e. for iterators more generally, not just class-type ones), as iterators are not required to be modifiable lvalues. See [Why can't I decrement std::array::end()?](https://stackoverflow.com/questions/48187786/why-cant-i-decrement-stdarrayend) for good discussion. So, basically, wrapping it in `std::prev()` is better style. – underscore_d Sep 22 '18 at 10:31
@mbrandalero *If* using predecrement is usable for a given iterator (see above comment), the decrement **must** be done before the assiignment, as that is how the entire world expects prefix operators to work, and any iterator not doing that would be dismally broken. – underscore_d Sep 22 '18 at 10:36

score 11 · Answer 3 · answered Apr 20 '10 at 20:15

11

With reverse iterators:

iter = (++n.rbegin()).base()

As a side note: this or Charles Bailey method have constant complexity while std::advance(iter, n.size() - 1); has linear complexity with list [since it has bidirectional iterators].

answered Apr 20 '10 at 20:15

Eugen Constantin Dinca

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Why "++n.rbegin()" ? To me that seems like a reverse iterator to the next to last element. Wouldn't "n.rbegin().base()" (without the "++") be the iterator to the last element ? – zentrunix Sep 11 '14 at 12:40
@JoséX. base() is not just shorthand for casting an reverse iterator into a forward iterator. Take a look at this response for more info about base: http://stackoverflow.com/a/16609146/153861 – Eugen Constantin Dinca Sep 22 '14 at 09:47

score 7 · Answer 4 · edited Nov 23 '11 at 05:25

7

Take the end() and go one backwards.

list <int>::iterator iter = n.end();
cout << *(--iter);

edited Nov 23 '11 at 05:25

cpx

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answered Apr 20 '10 at 19:58

Mitten.O

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Why do I have to go one backwards? Is there a dummy end node or something? – Tarion Jan 04 '18 at 17:09
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@Tarion: n.end() doesn't point at the last element, but right off the end of the list. – ratiotile Feb 07 '18 at 02:15

score 6 · Answer 5 · answered Jan 07 '13 at 06:11

6

std::list<int>::iterator iter = --n.end();
cout << *iter;

answered Jan 07 '13 at 06:11

B Faley

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Prefix operators on rvalue iterators will only work in some cases and are not required to, so you should instead use `std::prev()`; see https://stackoverflow.com/questions/2678175/iterator-to-last-element-in-stdlist#comment91853941_2678214 – underscore_d Sep 22 '18 at 10:41

score 1 · Answer 6 · edited Oct 10 '14 at 12:54

1

You could write your own functions to obtain a previous (and next) iterator from the given one (which I have used when I've needed "look-behind" and "look-ahead" with a std::list):

template <class Iter>
Iter previous(Iter it)
{
    return --it;
}

And then:

std::list<X>::iterator last = previous(li.end());

BTW, this might also be available in the boost library (next and prior).

edited Oct 10 '14 at 12:54

Jan Hudec

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answered Apr 20 '10 at 20:20

UncleBens

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This is available in C++0x as well (`std::next` and `std::prev`). – James McNellis Apr 20 '10 at 20:30

score 0 · Answer 7 · answered Nov 20 '12 at 23:54

0

list<int>n;
list<int>::reverse_iterator it;
int j;

for(j=1,it=n.rbegin();j<2;j++,it++)
cout<<*it;

answered Nov 20 '12 at 23:54

Alexandru Manescu

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Explaining the code you posted would make your answer even better. – Nov 21 '12 at 00:13

Iterator to last element in std::list

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