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Re: "Secret" operators
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From:
Andy_Bach
Date:
February 1, 2005 09:18
Subject:
Re: "Secret" operators
Message ID:
OFB4049FEF.EEDF7054-ON86256F9B.005E8C6E-86256F9B.005F0255@uscmail.uscourts.gov
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: andy_bach@wiwb.uscourts.gov
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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