Is it safe to use std::vector.erase(begin(), end()) or std::vector.erase(begin(), begin())?

Question

I want to process elements from a vector for some time. To optimize this I don't want to remove an item when it is processed and I remove all processed items at the end.

vector<Item*>::iterator it; for(it = items.begin(); it != items.end(); ++it) { DoSomething(*it); if(TimeIsUp()) { break; } } items.erase(items.begin(), it);

Is it safe to use erase when it == items.end()? In documentaion it is said that erase() will erase [first,last) and this should be safe but I want to be sure.

EDIT:

Is it safe to use std::vector.erase(begin(), begin())?

I strongly advise against using a vector of regular pointers. Either use a container specifically designed to hold pointers or use a vector of smart pointers. (A possible exception would be if the vector doesn't own the objects it points to.) — David Schwartz
– David Schwartz, Commented Dec 17, 2012 at 17:16
@Cristy: I don't think that is really a duplicate, the other Q was about a crash because of incrementing an iterator beyond end. It only incidentially contains erase and end, too. — Damon
– Damon, Commented Dec 17, 2012 at 17:19
@Damon Because it may not always be equal to items.end() — Praetorian
– Praetorian, Commented Dec 17, 2012 at 17:21
@Praetorian: Oh how stupid of me... you're right. I've read "remove all elements afterwards" instead of "removed all processed elements". — Damon
– Damon, Commented Dec 17, 2012 at 17:25

djechlin · Accepted Answer · 2012-12-17 17:11:34Z

5

Yes, this is correct - that's what the notation [first, last) means and end() points one past the end.

answered Dec 17, 2012 at 17:11

djechlin

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Comments

Joseph Mansfield · Accepted Answer · 2012-12-17 17:23:32Z

To answer your edit: yes, that's fine too. vec.erase(q1,q2) is defined as erasing "the elements in the range [q1, q2)." Here, both arguments are the same. [q1, q2) is defined as being a valid range of const iterators. A range is considered to be valid if there is a sequence of increments you can do to q1 to get to q2 - in this case, the sequence has no increments. In fact, the standard specifically defines an empty range:

A range [i,i) is an empty range

David Hammen · Accepted Answer · 2012-12-17 17:44:34Z

What that [first,last) means is everything between first and last, including first but not including last. It denotes a half-open set. Had they used [first,last], last would also be included; had they they used (first,last), first and last would be excluded.

There is a problem with your code. You want the element at it to be deleted if it isn't equal to end(). Instead of items.erase (items.begin(),it), you should be using

if (it != items.end()) { items.erase (items.begin(), it+1); } else { items.clear(); // same as items.erase (items.begin(), items.end()) }

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