zouz has asked for the wisdom of the Perl Monks concerning the following question:

Hi there, if someone can help me here with printing out the age in a shorter methode. here is what i am doing. 

print   "Alex children are

        $alex{children}->[1]->{age} 
and     $alex{children}->[0]->{age} years old\n";

print   "Alex is $alex{age} himself\n";

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is there any shorter method/ to do so? thank you in advance.

Replies are listed 'Best First'.
Re: short method to print age.
by wfsp (Abbot) on Mar 05, 2009 at 14:45 UTC
    One way could be to extract the childrens ages into an array using map before printing them. Perhaps also consider printf to format the output. 
    #!/usr/bin/perl

use warnings;
use strict;

my %alex = (
  age => 30,
  children => [
    {age => 2},
    {age => 5},
  ],
);

my @kids = map{$_->{age}} @{$alex{children}};

print   "Alex children are
        $kids[0] 
and     $kids[1] years old\n";
print   "Alex is $alex{age} himself\n\n";

# or

printf(
  qq{Alex's children are %d and %d years old\nAlex is %d himself\n},
  @kids, $alex{age}
);

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Re: short method to print age.
by ikegami (Patriarch) on Mar 05, 2009 at 16:50 UTC

    What if Alex has three children? Who cares about shorter if it doesn't produce the right output. 

    use strict;
use warnings;

sub plural($@) {
   my ($n, $s, $p) = @_;
   if    ($n == 0) { return "no $p";  }
   elsif ($n == 1) { return "one $s"; }
   else            { return "$n $p";  }
}

sub count_phrase {
   my $last = pop @_;
   return $last if !@_;
   return join(', ', @_), ' and ', $last;
}

sub print_person {
   my ($person) = @_;
   my $name     = $person->{name};
   my $age      = $person->{age};
   my $children = $person->{children};

   if ( @$children ) {
      my @chd_ages = map { $_->{age} } @$children;
      print("$name has ", count_phrase(@$children, 'child', 'children'
+), "\n");
      print("The are ", phrase_list(@chd_ages), " years old\n");
   } else {
      print("$name has no children\n");
   }

   print("$name is $age years old\n");
}

{
   my %person = (
      name     => 'Alex',
      age      => 30,
      children => [
         { age => 2 },
         { age => 5 },
      ],
   );

   print_person(\%alex);
}

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Re: short method to print age.
by Roy Johnson (Monsignor) on Mar 05, 2009 at 19:20 UTC
    This mimics what you do. 
    print "Alex children are \n\n        "
, join("\nand     ", map $_->{age}, reverse @{$alex{children}})
, "years old\n";

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    Obviously the reverse isn't necessary unless you really want it.
    Caution: Contents may have been coded under pressure.
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[d/l]
      Thank you all for the answers you guys offered. i think i will end up using this code. 
      #!/usr/bin/perl
use warnings;

%alex = (              
    'name' => 'Alex',
    'age'  => '45');
%john = (              
    'name' => 'John',
    'age'  => '20');
%chaddy = (            
    'name' => 'Chady',
    'age'  => '16');

@children = (\%john, \%chaddy);
$alex{'children'} = \@children;

print "Alex children are \n        "
, join("\nand     ", map $_->{age}, reverse @{$alex{children}})
, "\tyears old\n";
print   "Alex is $alex{age} himself\n";

      [download]
      instead of the one i mentioned above. much easier to understand i guess,thx Roy and thank you all once again.
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Re: short method to print age.
by locked_user sundialsvc4 (Abbot) on Mar 05, 2009 at 15:56 UTC

    You can certainly omit some of those -> symbols if you want to.

    For example, I'm sure you could write: “$alex->{children}[1]{age}”.

    You can also write “$$alex” as a shorthand for “$alex->”.

    Just remember to be clear and consistent. Use the directives “use strict; use warnings;” as a matter of inflexible routine.

      Saving one character there isn't going to affect the length of the expression. Furthermore, -> is usually much easier to read so your suggestion actually makes things worse.
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[d/l]

      Wasn't $$alex the original way to dereference? It was considered too confusing (or something) leading to the addition of the arrow operator.

      And you didn't even know bears could type.

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