I guessed you'd be having a go at improving it. :-)
I can always do with the exercise, Nigel. ;-)
Nice one. :-) Since the figure to be rounded is rendered positive
anyway, the "add 0.5 and div by 1" method can be used, which is one
less
step than 'it div 0.5 - it div 1'. Also 'character 2' seems to be
minutely faster than 'middle character'.
I tested the 'known position vs. middle' thing on the basis that
determining the middle item might involve some kind of internal
division. Whether or not that's the reason for the slight difference,
a direct positional reference (where the position is known) seems
preferable. (That said, 'middle' beats calculating the position of an
object's middle element, and then extracting the value.)
Interestingly, although multiplying an integer by itself is nearly
twice
as fast as squaring it (!), the effect was for some reason almost
exactly cancelled out by the two tells in my script. :-\
Yeah. Even a tell takes a toll, I guess. ;-)
My effort did better than your original with one of my favourite
test numbers, -321.255, but they were both wrong with, for instance, -
321.575. Your later version seems to be very robust so far. :)
Encouraging. And with a little shuffling, we can take advantage of
both a multiplication and a single tell:
------------------
on currency_format for v
tell (v mod 1 as text) * 100 to (v div 1 as text) & (0.0 as text)'s
character 2 & ((it * it) ^ 0.5 + 0.5) div 1
end currency_format
currency_format for -321.255
--> "-321,26" / "-321.26" (* depending on local currency format *)
