Re: why variable inside submodule did not change value

[itshardtogetone]

Thanks Rob and members,
What do you mean by "but since you have declared a subroutine that 
references the variable...."
Thank You.
Ed

----- Original Message ----- 
From: "Rob Dixon" <[EMAIL PROTECTED]>
To: <beginners@perl.org>
Cc: <[EMAIL PROTECTED]>
Sent: Sunday, March 23, 2008 9:06 AM
Subject: Re: why variable inside submodule did not change value


[EMAIL PROTECTED] wrote:
Hi,
With regards to the script below, can somebody explain to me why
testcounter after counting 1,2,3, remains at 3 all the time. I thought
it would have been reset to zero by the outer while loop.
Thanks

#!c:\perl\bin\perl
use strict;
my $anything = 5;
my $counter =0;

while ($anything > $counter){
     my $testing = 3;
     my $testcounter = 0;
    $counter += 1;

     while ($testing > $testcounter ){
        $testcounter += 1;
        test() if (1);

        sub test {
          print "$testcounter.\n\n";
        }
     }
}

The statement

 my $testcounter = 0;

will create the variable and initialize its value to zero. Ordinarily
the variable will be released at the end of the loop and recreated (and
initialized) when the loop starts again, but since you have declared a
subroutine that references the variable, its reference count never goes
to zero and it is never released, and so never reinitialized. This
construct is called a closure, and can be useful if you understand it,
but here it is simply causing problems. If you always declare your
subroutines at file scope then you will avoid this. Look at the program
below to see how.

HTH,

Rob


use strict;
use warnings;

my $anything = 5;
my $counter = 0;

while ($anything > $counter) {

 my $testing = 3;
 my $testcounter = 0;

 $counter += 1;

 while ($testing > $testcounter ) {
   $testcounter += 1;
   test($testcounter);
 }
}

sub test {
 my $n = shift;
 print "$n.\n\n";
}




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