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

[Brian Milby]

Here's an image of the stack in my fork of the repo:
https://github.com/bwmilby/alloffsets/blob/bwm/bwm/stack_allOffsets_card_id_1018.png


On Sun, Nov 4, 2018 at 10:07 PM Brian Milby <brian at milby7.com> wrote:

> I’m working on an update to the stack now. Moving buttons to the left side
> to make it easier to add more.
>
> Thanks,
> Brian
> On Nov 4, 2018, 10:02 PM -0600, Mark Wieder via use-livecode <
> use-livecode at lists.runrev.com>, wrote:
>
> On 11/4/18 4:45 PM, Brian Milby via use-livecode wrote:
>
> 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.
>
>
> Nice. I tried to get tricky and replace that 'replace with' loop with a
> 'repeat for each' loop, but ended up about 20% slower. Not at all what I
> expected.
>
> --
> Mark Wieder
> ahsoftware at gmail.com
>
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