What does $$ in perl actually return?

3
votes

I'm new to Perl, learning from the beginning. I've read that $$ returns

The pid of the Perl process running this script.

as per below link,

http://www.tutorialspoint.com/perl/perl_special_variables.htm

I've the following Perl Script which is being executed on a Windows machine,

sub test
{
    my ($surrogate_name) = @_;
    if((defined $surrogate_name) && ($surrogate_name == 1))
    {
        print "Done!"."\n";
    }
    return $surrogate_name;
}

sub t
{
    my ($surrogate_name) = @_;
    my $record;
    my $surrogate_value;
    $record = &test($surrogate_name);
    print $record."\n";
    if ($record)
    {
        print "B"."\n";
        $surrogate_value = $$record{$surrogate_name};
    }
    print $surrogate_value."\n";
    print "A"."\n";
    return $surrogate_value;
}

&t(1);

In this code, I observed everything is printed except the value of $surrogate_value.

Please clarify me what does $$ actually mean and why it's not returning anything in my script.

Thanks in advance...

3
perl
always use strict and warnings. It looks like the $$ here is either a typo, or a scalar dereference. - Patrick J. S.
What tutorial are you reading that tells you to use an ampersand & to call subroutines, as in &test($surrogate_name)? Whatever it is, you should stop reading it as it is around twenty years out of date! Just test($surrogate_name) is correct. - Borodin
Please don't use tutorials from tutorialspoint.com. They are terrible. Far better to look at learn.perl.org and the Perl Tutorial Hub. - Dave Cross

3 Answers

12
votes

Yes $$ returns the process id of currently running script.

but in your case $surrogate_value = $$record{$surrogate_name} which is completely different concept that is a dereferencing.

for example $$ used along with a variable name to dereference it.

my $a = 10; #declaring and initializing a variable.
my $b = \$a; #taking scalar value reference
print $$b;  #now we are dereferencing it using $$ since it is scalar reference it will print 10

my %hashNew = ("1" => "USA", "2" => "INDIA"); #declaring a hash
my $ref = \%hashNew; #taking reference to hash
print $$ref{2}; #this will print INDIA we derefer to hash here

For more understanding read referencing and dereferencing in perl.

1
votes

Normally the $$ is used to print the current process ID.

print $$;

But the $ is having another work, which for dereferencing the scalar variable. For example:

use strict;
use warnings;

my $Input = 5;
my $Input_Refer = \$Input;
print "$Input\n";
print $$Input_Refer;

In the above example, we are have two scalar variable, 1. $Input, which contains 5 as value. 2. $Input_Refer, which contains the reference of an $Input variable as value. So If we print the $Input, It will give the value as 5, If we print the $Input_Refer, It will print the memory address of $Input. So we have to dereference it, For that we have to use another $ like this,

print $$Input_Refer;

It will give the output as 5.

1
votes

The variable $$ is, as you've written, PID of current process:

print $$ . "\n";

What you've written in your script is $$record{$surrogate_name}, which means accessing an element of a hash and is equivalent to $record->{$surrogate_name}.

Other from that, $$name usually means dereferencing a reference to a scalar. For example, if you have:

my $x = 1;    # Scalar
my $y = \$x;  # Scalar, value of which is a reference to another scalar (address)
my $z = $$y;  # Dereference the reference, obtaining value of $x

It's equivalent to following operations on pointers in C:

int  x = 1;
int *y = &x;
int  z = *y;

Read more on this here.