Re: "Secret" operators

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"))."·
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