How to find the offset of the last instance of a repeating character in a string? (Geoff Canyon)

[Brian Milby]

My updated solution always looks for overlap but if none are found it uses optimized versions of the search (private functions instead of inside the main function). I special case for no overlap and a single overlap in the delimiter. It is about the same speed as Geoff’s.

Thanks,
Brian
On Nov 4, 2018, 6:34 PM -0600, Mark Wieder via use-livecode <use-livecode at lists.runrev.com>, wrote:
> On 11/4/18 10:40 AM, Geoff Canyon via use-livecode wrote:
> > I also added a "with overlaps" option.
>
> My problem with the pWithOverlaps parameter is that is requires a priori
> knowledge of the data being consumed. If you already know there are
> overlaps then you'd set the parameter to true. If you don't know whether
> or not there are overlaps, then you'd need to set it to true so you
> don't miss anything (aside, of course, for the trivial case where you
> don't care whether or not there are overlaps - is there a use case for
> this?).
>
> The only time you would set it to false is after you've already
> determined that there are no overlaps, and the time spent on that would
> probably more than offset the extra processing in the function.
>
> --
> Mark Wieder
> ahsoftware at gmail.com
>
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