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Iterator to last element in std::list

开发者 https://www.devze.com 2022-12-27 19:03 出处:网络
#include <list> using std::list; int main() { list <int> n; n.push_back(1); n.push_back(2); n.push_back(3);
#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 w开发者_如何学JAVAay to have an iter to the last element in list?


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;


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();


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].


Take the end() and go one backwards.

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


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


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).


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

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