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

[Brian Milby]

Logic matches my solution.  I also validated my solution using just the
offset function.  Speed hit for with overlap is similar.  One possible
optimization:

put kList is not empty into pWithOverlaps

If with overlaps was requested but the source delimiter did not contain any
overlaps, then the extra loops are skipped.

Adding a character to the end is clever.  I'll need to incorporate that and
see what it does to my method.

My take on the code updates is here:
https://github.com/bwmilby/alloffsets/blob/bwm/bwm/allOffsets_Scripts/stack_allOffsets_button_id_1026.livecodescript

Stack and index of scripts here:
https://github.com/bwmilby/alloffsets/tree/bwm/bwm

On Sun, Nov 4, 2018 at 12:42 PM Geoff Canyon via use-livecode <
use-livecode at lists.runrev.com> wrote:

> Alex, good catch! The code below and at
> https://github.com/gcanyon/alloffsets now puts a stop character after the
> string to prevent the error you found. I also added a "with overlaps"
> option. I think this is correct, and about as efficient as possible, but
> thanks to anyone who finds a bug or a faster way.
>
> gc
>
>
> function allOffsets D,S,pCase,pWithOverlaps
>    -- returns a comma-delimited list of the offsets of D in S
>    set the caseSensitive to pCase is true
>    put length(D) into dLength
>    put numtochar(chartonum(char -1 of D) mod 2 + 1) after S
>    if pWithOverlaps then
>       repeat with i = 1 to dLength - 1
>          if not (char i + 1 to -1 of D is char 1 to dLength - i of D) then
> next repeat
>          put char -i to -1 of D into OV[i]
>          put i & cr after kList
>       end repeat
>    end if
>    set the itemDel to D
>    put 1 - dLength into C
>    if pWithOverlaps then
>       repeat for each item i in S
>          repeat for each line K in kList
>             if char 1 to K of (i & D) is OV[K] then put (C + K),"" after R
>          end repeat
>          add length(i) + dLength to C
>          put C,"" after R
>       end repeat
>    else
>       repeat for each item i in S
>          add length(i) + dLength to C
>          put C,"" after R
>       end repeat
>    end if
>    set the itemDel to comma
>    repeat until item 1 of R > 0
>       delete item 1 of R
>    end repeat
>    delete item -1 of R
>    if R is empty then return 0 else return char 1 to -2 of R
> end allOffsets
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