Re: Re: quick question about hash element access

added to your code so it can take a variable length identifier, and accept the hashref.


my %test_hash;
$test_hash{'a'}{'b'}{'c'}{'d'} = "Yay";
print get_id(\%test_hash, "a.b.c.d"), "\n";

sub get_id
{
    my $thingy = shift;
    my @keys = split /\./, shift;
    
    foreach my $key ( @keys )
    {
        $thingy =  $$thingy{$key};
    }
    return $thingy;
    
}
[download]

Comment on Re: Re: quick question about hash element access Download Code

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Re: Re: Re: quick question about hash element access by Nkuvu (Priest) on Jun 27, 2003 at 21:45 UTC
Interesting. But I have a question (not being totally fluent in references). What exactly does the `$thingy = $$thingy{$key}` line do? My guess is that it assigns a scalar value referred to by the reference which is contained in `$thingy{$key}`. But I'm not even sure that sentence makes sense, much less if it's accurate. :)	[reply] [d/l] [select]
Re: Re: Re: Re: quick question about hash element access by LanceDeeply (Chaplain) on Jun 27, 2003 at 22:32 UTC
you are correct sir- we are assigning `$thingy` to the value of whatever is refered to by `$thingy{'a'}`, in this case, it's the next hashref. this assignment is just repeated a few times till we're all out of keys. what's bad about this is that it assumes the caller will pass in a valid path. i guess error checking will be left up to OP...	[reply] [d/l] [select]
Re: Re: Re: Re: quick question about hash element access by argggh (Monk) on Jun 28, 2003 at 14:06 UTC
`$$thingy{$key}` is equal to `$thingy->{$key}`, so it returns the value corresponding to the key `$key` in the hash pointed to by the reference `$thingy`.	[reply] [d/l] [select]


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