If you have a linked list, you can quickly swap elements by switching around the appropriate pointers. But if you have an std::list and apply std::swap, (which happens to be the only way I know how to swap elements in a container), it will swap elements by performing a few copy operations. Is there a way I can get an std::list to perform a swap using pointers instead?

You could probably do it with list::splice(). However, if the performance of the swap behavior bothers you, you may want to consider using a list of (smart) pointers instead of storing the elements by value.

Quote:Original post by fpsgamer
If you have a linked list, you can quickly swap elements by switching around the appropriate pointers.

But if you have an std::list and apply std::swap, (which happens to be the only way I know how to swap elements in a container), it will swap elements by performing a few copy operations.

Is there a way I can get an std::list to perform a swap using pointers instead?

By putting pointers inside the container in the first place. (std::list< X* >)

Swap is swap - there's nothing you can do about it, you have no control over how things are stored internally, or whether you may just swap pointers.