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

[Geoff Canyon]

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