From the keyboard of [EMAIL PROTECTED] [01.02.05,11:15]:
> Can I get that just a little slower?
>
> $b = () = /u/g;
>
> is the same as:
>
> @a = /u/g;
> $b = @a;
it's not the same.
perl -le '$_="foo"; print $b =()= /o/g'
2
perl -le '$_="foo"; print @b =()= /o/g'
perl -le '$_="foo"; print /o/g'
oo
The brackets in =()= means something like 'here would have been n elements,
were you interested'. There's no list, only list context.
-gg-
> I understand what happens, but it appears to be assigning to an empty list
> - is that filling up the list, so to speak? Or is it just that it makes
> the 'result' of /u/g assign in array/list context and then that, assigned
> in scalar context to $b gives the list/array count.
>
> I tried to explain this once and I had to resort to 'and then <mumble
> mumble> and in scalar context, we get the count of the elements in $b!'
>
> a
>
> Andy Bach, Sys. Mangler
> Internet: [EMAIL PROTECTED]
> VOICE: (608) 261-5738 FAX 264-5932
>
> "Bugs happen. A bug is a test case you haven't written yet."
> Mark Pilgrim
>
--
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/\_¯/(q /
---------------------------- \__(m.====·.(_("always off the crowd"))."·
");sub _{s,/,($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e,e && print}
