Re: "Secret" operators

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."
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