Re: Passing by reference from within a sub

[RichardK]

Isn't this just an ordering issue ?

The compiler hasn't seen the definition for sub_c when it compiles sub_b, but it has by the time it gets to the stand alone call of sub_c, hence the differing behaviour. So if you swap the order of your functions it will do what you want.

use strict;
use Data::Dumper;

sub sub_c (\%)
{
    my ($hash1c) = @_;

    print "sub_c arguments:" . Dumper (@_);
}

sub sub_b ()
{
    my %hash1b = ();
    $hash1b{"key1"} = "1";

    print "calling sub_c from sub_b\n";
    sub_c (%hash1b);
}


my %hash1a = ();
$hash1a{"key1"} = "1";

print "calling sub_b\n";
sub_b ();

print "\n";
print "calling sub_c directly\n";
sub_c (%hash1a);

[arvid]

Thank you Richard for saving my sanity. It works like expected now.