<algorithm>: search_n returns first when count<=0

[andreyfe1]

Hi,
It was found that search_n returns first when count<=0.
However, the C++ standard says (N4659 draft, the last draft before C++17):

Returns: The first iterator i in the range [first, last-count) such that for every non-negative integer n less than count the following corresponding conditions hold: (i + n) == value, pred((i+ n),value) != false. Returns last if no such iterator is found.

It seems that if n>=0 and count <=0 we don't have a chance to apply a predicate. So, we won't find "such iterator". According to this logic last should be returned. Am I right?
I'm just curious why does search_n implementation return first?

[StephanTLavavej]

Returning first is correct because "for every non-negative integer n less than count" is vacuously true when count <= 0.

The Standardese here could be clarified along the lines of
N4830 25.6.7 [alg.generate]/1 "Let N be max(0, n)". I'm tagging this as LWG-issue-needed, but I'd actually suggest filing an editorial issue since there's no behavioral change or restriction involved here.

[andreyfe1]

Thanks @StephanTLavavej . It's pretty interesting. I wouldn't say that "for every non-negative integer n less than count" leads to first returning. Anyway, it should be clarified in the standard. Let me submit an editorial issue.

[CaseyCarter]

Now that an editorial issue has been submitted for the C++ working draft, this is in WG21's hands. There's nothing actionable remaining here.