Can I get that just a little slower? $b = () = /u/g;
is the same as: @a = /u/g; $b = @a; 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
