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Javier Guerra a écrit :
On Thursday 21 December 2006 8:21 am, Roberto Ierusalimschy wrote:
Is there a more elegant way?
No :(  (At least you can be confident that the generated opcode is
as good as for a primitive continue ;)

i guess both metalua's and token filter's wizards will soon present their solutions.

but this is not the first time this has been discussed. i beleive there was a serious semantic problem; but i can't remember it now

i'd really wish there was a 'continue' statement, in fact i've been using deep if's, for me the 'break out of repeat' construct is an improvement!!

Yes, I frown as it was a kind of Faq, but since Leo came with a good solution, that is even marked as "efficient" by Roberto himself, no problem!
I will add this to my toolbox... ;-)

for i = 0, 9 do repeat
 if math.mod(i, 2) == 0 then break end
until true end

It has a look of "special loop with continue" syntax... :-)

Philippe Lhoste
--  (near) Paris -- France
--  --  --  --  --  --  --  --  --  --  --  --  --  --