copying vector to list

Hello

I have a std::vector<std::string> with some elements..
I would like to copy last X elements to another std::list<std::string> and remove them from std::vector.

What is the easiest/best way to do this?

I know I should use std::copy and do interation, but not sure about the second step.

Thanks a lot for help!

Regards

iterators would work well I think.

Code:

std::vector<std::string> myvector;
std::list<std::string> mylist;

std::vector<std::string>::iterator iter;
for ( iter = myvector.begin(); iter != myvector.end(); ++iter )
{
  mylist.push_back((*iter));
 myvector.erase(iter);
}

that will copy the entire vector, and erase each element as it goes. you could modify that to take only certain elements.
just put + however many elements you want to skip after the .begin()

Code:

iter = myvector.begin()+SkipAmount;

Hopefully that helped.

EDIT:
If your wanted to always only take so many elements, then maybe a reverse_iterator would work better.

Thanks for reply but arent there any simpler/better methods (maybe like):

std::copy(vector.begin() + 10, vector.end(), ..);
std::del(vector.begin() + 10, vector.end(), ..);

Yea. I am a little weird though. I like writing my own solutions for such problems. Either way you do it will be about the same amount of code.
might try looking at the reference on http://www.cplusplus.com
The reference there is thorough, and explains things well.

Code:

std::copy(vec.begin(), vec.end(), std::back_inserter(list));
vec.clear();

Use the erase-remove idiom for the 2nd step if you only have access to vector iterators; if you have access to the vector object itself, use the vector.erase() function.

Originally Posted by cpjust

Use the erase-remove idiom for the 2nd step.

I prefer to assume the container will be reused some time in the near future -- unless it's possible to show otherwise. If it will be reused, no point releasing its storage. If it will not be reused, I question why it does not immediately go out of scope at that point.

EDIT: Nevermind that whole thing. I was looking at the swap-trick part of the page you linked. Sorry.

Originally Posted by brewbuck

I prefer to assume the container will be reused some time in the near future -- unless it's possible to show otherwise. If it will be reused, no point releasing its storage. If it will not be reused, I question why it does not immediately go out of scope at that point.

I think l2u needed to remove some of the vector, not all of it.

I would like to copy last X elements to another std::list<std::string> and remove them from std::vector.

Originally Posted by cpjust

I think l2u needed to remove some of the vector, not all of it.

Exactly..

I want to copy and remove some objects (for instance last 10 in the vector), not all of them.

I would suggest taking a look at list's insert method (3rd overload in that reference) and vector's erase method (again the one that takes a range).

To copy x elements, creating a new list:

Code:

std::vector<int>::iterator from = vec.end() - x;
std::list<int> result(from, vec.end());
from.erase(from, vec.end());

If the list exists, use insert() instead.

Originally Posted by CornedBee

To copy x elements, creating a new list:

Code:

std::vector<int>::iterator from = vec.end() - x;
std::list<int> result(from, vec.end());
from.erase(from, vec.end());

If the list exists, use insert() instead.

I always forget about the range setting constructor.

Originally Posted by CornedBee

To copy x elements, creating a new list:

Code:

std::vector<int>::iterator from = vec.end() - x;
std::list<int> result(from, vec.end());
from.erase(from, vec.end());

If the list exists, use insert() instead.

What type should x be?
Is it int x safe?

Technically, it should be std::iterator_traits<std::vector<int>::iterator>:: difference_type. In all likelihood, that's a typedef for std:trdiff_t, i.e. a signed integer of platform word size. 32 bits on 32-bit platforms, 64 on 64-bitters. Practically, though, unless you actually have a vector larger than 2 billion elements, int is fine.