Ariel Scolnicov wrote:
> 
> 
> So how do I make C<foo> into an array in the first place?  Well, I say
> something like C<foo = (1,2,3)>.  But wait -- that's ambiguous!  Is
> C<foo> now a copy of the list (1,2,3) (in which case it's an array),
> or is it a reference to (1,2,3) (in which case it's a scalar)?  In the
> first case, C<foo = bar> (where C<bar> is an array, assuming I could
> ever assign an array to a variable) would perform a copy, while in the
> second it would make the two variables refer to the same thing.

Actually, if you wanted an array, you'd say C<foo = [1,2,3]>.  Very
unambiguous.

> 
> --
> Ariel Scolnicov        |"GCAAGAATTGAACTGTAG"            | [EMAIL PROTECTED]
> Compugen Ltd.          |Tel: +972-2-6795059 (Jerusalem) \ We recycle all our Hz
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-- 
David Corbin            
Mach Turtle Technologies, Inc.
http://www.machturtle.com
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