in reply to (tye)Re: Why no stay shared warning?
in thread Why no stay shared warning?

It seems that my $foo only creates a place for sub up to store values between calls. Why does initializing it to 20 not have an effect?
Thanks 
use strict; use diagnostics;
up();
up();
print "getfoo=",get();
{
my $foo = 20;
sub up {
    print "subup=$foo\n";
    $foo++;
sub get {$foo}
}
}

prints:

subup=
subup=1
getfoo=2

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Re: Re: (tye)Re: Why no stay shared warning?
by japhy (Canon) on May 05, 2001 at 20:01 UTC
    The reason is because $foo = 20 doesn't occur until after you've called all those functions.

    DECLARATION occurs at compile-time. ASSIGNMENT happens at run-time. Different things. my $foo = 20 has a compile-time effect (my) and a run-time effect ($foo = 20).

    For your desired action, you'd need something like: 
    my $foo;
BEGIN { $foo = 20 }

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    japhy -- Perl and Regex Hacker
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