in reply to Re: Sub hash param by reference
in thread Sub hash param by reference

Thank you Marshall, my $ref = \$hash{id2} is exactly what I am trying to look for.

Your given example is perfect. I'm just curious, when calling sub modify_anon_hash_value ($ref) is it possible to do it like modify_anon_hash_value (\$hash{id2}) ? without need to do my $ref = \$hash{id2}.

I have test run it with 

modify_anon_hash_value (\$hash{id2});

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The result return correctly without any error. Just to confirm if this way of coding is allowed or it may be a bad way to do it.

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Re^3: Sub hash param by reference
by Marshall (Canon) on Jul 08, 2016 at 07:06 UTC
    Yes, you will be fine without creating an intermediate scalar. Using this method (modify_anon_hash_value (\$hash{id2});), I see right away in the callers code that a modification to $hash{id2} is what is intended.

    What is sent to the sub are aliases. It is possible to "hide" this reference creation in the sub. For fun, I coded the routine "confusing" below. The caller wouldn't expect that to happen, so I wouldn't do that.

    #!/usr/bin/perl
use strict;
use warnings;
use Data::Dump qw(pp);
$!=1;

my %hash;
$hash{id1} = 001;
$hash{id2} = 002;
print "initial hash: ", pp (\%hash), "\n";

my $hash_ref = \%hash;

modify_anon_hash_value (\$hash{id2});
print "mod by sub   : ", pp (\%hash), "\n";

confusing ($hash{id2});
print "mod by confusing   : ", pp (\%hash), "\n";

sub modify_anon_hash_value  #sub doesn't know which
{                           #key's value that it is
   my $ref = shift;         #modifying
   $$ref = 5;
}

sub confusing  #a confusing mess
{ 
   my $ref = \(shift);    #same as $ref = \$_[0]
   $$ref=9;               # @_ has aliases, not a copy
}

__END__
initial hash: { id1 => 1, id2 => 2 }
mod by sub   : { id1 => 1, id2 => 5 }
mod by confusing   : { id1 => 1, id2 => 9 }

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      Thank you again Marshall. The confusing ($hash{id2}); is really confusing, esp my $ref = \(shift);. I will keep that in mind and avoid doing so.

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