Constant time std::list resize to a given last element?

Question

Resizing a std::list is linear time since it needs to find the n th element first.

But is it possible to point to an element (using iterator) and resize in constant time by declaring that the pointed-to element is the new last element?

It would be something like: make_last(const_iterator pos) .

Answer 1

Unfortunately, in C++11 general std::list splicing is O(n), in order to get O(1) size checking.

Otherwise you could have moved the undesired items to another list.

But even that's just a half solution, since the items have to be destroyed at some point. However, if constant time splicing was guaranteed this would make a temporary resize O(1).

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A practical solution is to use eg std::vector instead of std::list .

Due to necessity of destroying it doesn't give you O(1) resizing either, except when the resizing discards a constant number of items at the end, but it avoids all that complexity.

With the decision to make std::list::size a constant time operation, I do not know of any remaining reason to use std::list for anything.

Answer 2

In C++03 you use erase on the first element you don't want - constant time iff the optimiser realises the elements don't need destruction.

In C++11 you're stuffed... it's obliged to work out the new size as Alf explains.