best way that I know of is to use addAt:
cA = listA.count
cB = listB.count
if cA < cB then
listC = listB.duplicate()
repeat with i = aC down to 1
listC.addAt(1, listA[i])
end repeat
else
listC = listA.duplicate()
repeat with i = aC down to 1
listC.addAt(1, listB[i])
end repeat
end if
listC will contain the shorter list placed in front of the longer.
similarly, one could also use append & place the shorter list's
elements onto the end.
note: string converting of lists is really slow.
hth
-Buzz
At 2:39 PM -0500 4/10/02, you wrote:
>so I've got 2 prop lists; I'd like to append one to the other witout
>having to do the 1 to count of the second....there's no other way, right?
>
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