Reputation: 4555
See related questions on past-the-end iterator invalidation: this, this.
This is more a question of design, namely, is there (in STL or elsewhere) such concept as past-the-end iterator "revalidation"?
What I mean by this, and use case: suppose an algorithm needs to "tail" a container (such as a queue). It traverses the container until end()
is reached, then pauses; independently from this, another part of the program enqueues more items in the queue. How is it possible for the algorithm to (EDIT) efficiently tell, "have more items been enqueued" while holding the previously past-the-end iterator (call it tailIt
)? (this would imply it is able to check if tailIt == container.end()
still, and if that is false, conclude tailIt
is now valid and points to the first element that was inserted).
Please don't dismiss the question as "no, there isn't" - I'm looking to form judgment around how to design some logic in an idiomatic way, and have many options (in fact the iterators in question are to a hand-built data structure for which I can provide this property - end() revalidation - but I would like to judge if it is a good idea).
EDIT: made it clear we have the iterator tailIt
and a reference to container
. A trivial workaround for what I'm trying to do is, also remember count
:= how many items you processed, and then check is container.size() == count
still, and if not, seek to container[count]
and continue processing from there. This comes with many disadvantages (extra state, assumption container doesn't pop from the front (!), random-access for efficient seeking).
Upvotes: 5
Views: 292
Reputation: 385104
Not in general. Here are some issues with your idea:
Put all these together and it is clear that the end iterator simply cannot be used this way as the iterator interface is currently designed. Iterators refer to data in a range, not to a container; it stands to reason, then, that they hold no information about a container, and if the container causes the range to change there's no entity that the iterator knows about that it can ask to find this out.
Is the described logic possible to create? Certainly! But with a different iterator interface (and support from the container). You could wrap the container in your own class type to do this. However, I advise against making things that look like standard iterators but behave differently; this will be very confusing.
Instead, encapsulate the container and provide your own wrapper function that can directly perform whatever post-enqueuement action you feel you need. You shouldn't need to watch the state of the end iterator to achieve your goal.
Upvotes: 5
Reputation: 154
In the case for a std::queue, no there isn't (heh). Not because the iterators for a queue get invalidated once something is pushed, but because a queue doesn't have any iterators at all.
As for other iterator types, most (or any of them) of them don't require a reference to the container holder (the managing object containing all the info about the underlying data). Which is an trade-off for efficiency over flexibility. (I quickly checked the implementation of gcc's std::vector::iterator)
It is possible to write an implementation for an iterator type that keeps a reference to the holder during its lifetime, that way the iterators never have to be invalidated! (unless the holder is std::move'd)
Now to throw in my professional opinion, I wouldn't mind seeing a safe_iterator/flex_iterator for cases where the iterator normally would be invalidated during iterations.
Possible user interface:
for (auto v : make_flex_iterator(my_vector)) {
if (some_outside_condition()) {
// Normally the vector would be invalidated at this point
// (only if resized, but you should always assume a resize)
my_vector.push_back("hello world!");
}
}
Literally revalidating iterators might be too complex to build for it's use case (I wouldn't know where to begin), but designing an iterator which simply never invalidates is quite trivial, with only as much overhead as a for (size_t i = 0; i < c.size(); i++);
loop.
But with that said, I cannot assure you how well the compiler will optimize, like unrolling loops, with these iterators. I do assume it will still do quite a good job.
Upvotes: 1