in reply to Re^2: Sub hash param by reference
in thread Sub hash param by reference

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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Re^4: Sub hash param by reference
by hankcoder (Scribe) on Jul 08, 2016 at 13:05 UTC

    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.