Thanks a lot for your good explaining.I have been somewhat clear about it.
I'll do more reading just as you said.:)
Angel Flower
From: Perrin Harkins <[EMAIL PROTECTED]>
To: angel flower <[EMAIL PROTECTED]>
CC: "modperl@perl.apache.org" <modperl@perl.apache.org>
Subject: Re: a new mod_perl problem
Date: Wed, 19 Oct 2005 12:03:58 -0400
[ Please keep your questions on the list. ]
On Wed, 2005-10-19 at 12:39 +0800, angel flower wrote:
> hi,perrin,
> Can you tell me what meaning of this sentence:
>
> Making a sub that refers to a lexical variable declared outside of its
> scope will ALWAYS create a closure.
>
> Why this happen? And what's a closure? Thanks more.
This is kind of a big question, and you'll need to do some reading. I
recommend the section on closures in Programming Perl, and there's a
"What's a Closure?" entry in the perlfaq7 man page, and more in the
perlref man page.
The problem with all of these is that they only talk about closures with
anonymous subs, and closures really have nothing to do with anonymous
subs.
Here's a simple example:
package Private;
my $hidden = 0;
sub increment_hidden {
$hidden++;
}
sub get_hidden {
return $hidden;
}
1;
# in some other piece of code....
use Private;
print Private::get_hidden(); # prints 0
Private::increment_hidden();
print Private::get_hidden(); # prints 1
print $Private::hidden; # undefined variable
You might have expected that $hidden would get reset to undef when it
goes out of scope in the Private module. Instead, the two subs that
refer to it become closures, and they get a private copy of the
variable. Nothing else can access that variable at this point, but it
does persist as if it were a global for those two subroutines.
That's the short answer. Read some more docs, and remember that named
subs can be closures.
- Perrin
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