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From: PerlFAQ Server on 29 Mar 2010 12:00
This is an excerpt from the latest version perlfaq7.pod, which
comes with the standard Perl distribution. These postings aim to
reduce the number of repeated questions as well as allow the community
to review and update the answers. The latest version of the complete
perlfaq is at http://faq.perl.org .

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7.16: How do I create a static variable?

(contributed by brian d foy)

In Perl 5.10, declare the variable with "state". The "state" declaration
creates the lexical variable that persists between calls to the
subroutine:

sub counter { state $count = 1; $counter++ }

You can fake a static variable by using a lexical variable which goes
out of scope. In this example, you define the subroutine "counter", and
it uses the lexical variable $count. Since you wrap this in a BEGIN
block, $count is defined at compile-time, but also goes out of scope at
the end of the BEGIN block. The BEGIN block also ensures that the
subroutine and the value it uses is defined at compile-time so the
subroutine is ready to use just like any other subroutine, and you can
put this code in the same place as other subroutines in the program text
(i.e. at the end of the code, typically). The subroutine "counter" still
has a reference to the data, and is the only way you can access the
value (and each time you do, you increment the value). The data in chunk
of memory defined by $count is private to "counter".

BEGIN {
my $count = 1;
sub counter { $count++ }
}

my $start = counter();

.... # code that calls counter();

my $end = counter();

In the previous example, you created a function-private variable because
only one function remembered its reference. You could define multiple
functions while the variable is in scope, and each function can share
the "private" variable. It's not really "static" because you can access
it outside the function while the lexical variable is in scope, and even
create references to it. In this example, "increment_count" and
"return_count" share the variable. One function adds to the value and
the other simply returns the value. They can both access $count, and
since it has gone out of scope, there is no other way to access it.

BEGIN {
my $count = 1;
sub increment_count { $count++ }
sub return_count { $count }
}

To declare a file-private variable, you still use a lexical variable. A
file is also a scope, so a lexical variable defined in the file cannot
be seen from any other file.

See "Persistent Private Variables" in perlsub for more information. The
discussion of closures in perlref may help you even though we did not
use anonymous subroutines in this answer. See "Persistent Private
Variables" in perlsub for details.



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