Dereferencing hash reference passed to a subroutine

Question

I'm working through some sample code, and I'm trying to figure out why it works with the brackets around $args. Without it, I don't get the values.

sub random_dice{
  my ($args) = @_;
  my $number_of_rolls = $args->{number_of_rolls} || 6;
  ...
}

# I don't understand why it works with the brackets around $args
my $r = random_dice({number_of_rolls=>5});

See this answer: http://stackoverflow.com/questions/10031455/perl-using-my-with-parentheses-and-only-one-variable — squiguy, Apr 02 '13 at 20:28
read the tutorial on references: [perldoc perlreftut](http://p3rl.org/reftut) — Joel Berger, Apr 02 '13 at 20:51

score 5 · Accepted Answer · answered Apr 02 '13 at 20:27

It works because you are passing an anonymous hash to your random_dice subroutine.

my ($args) = @_; # sets $args as element of @_. Not as an array

$args is now a reference to a hash

$args = {
   number_of_rolls => 5
};

This is generally used as a method to have named parameters in Perl

score 3 · Answer 2 · answered Apr 02 '13 at 20:30

Like this:

my $args = @_;

the assignment is made in scalar context, so $args is assigned the value 1 (the number of elements in the array).

But like this:

my ($args) = @_;

the assignment is made in list context. The values from the array on the right side are unpacked and assigned to elements of the array on the left side.

ikegami · Answer 3 · 2013-04-02T20:40:25.077

2

If you want to use the simpler

my $r = random_dice( number_of_rolls => 5 );

the sub would have to be changed to

sub random_dice{
  my %args = @_;
  my $number_of_rolls = $args{number_of_rolls} || 6;
  ...
}

edited Apr 02 '13 at 20:40

answered Apr 02 '13 at 20:34

ikegami

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Dereferencing hash reference passed to a subroutine

3 Answers3