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 $b!' > > a > > Andy Bach, Sys. Mangler > Internet: [EMAIL PROTECTED] > VOICE: (608) 261-5738 FAX 264-5932 > > "Bugs happen. A bug is a test case you haven't written yet." > Mark Pilgrim > -- _($_=" "x(1<<5)."?\n".q·/)Oo. G°\ / /\_¯/(q / ---------------------------- \__(m.====·.(_("always off the crowd"))."· ");sub _{s,/,($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e,e && print}