Re: Correct data structure / sort method to use

Niall Macpherson <niall.macpherson@xxxxxxxxxxxxxxxxx> wrote in comp.lang.perl.misc:
> Anno Siegel wrote:
>
>
> >
> > Then make the hash(ref) a parameter of the function instead of (or in
> > addition to) the keys. Untested:
> >
> > sub sort_by_used {
> > my $hr = shift;
> > sort { $hr->{ $a}->{ used} <=> $hr->{ $b}->{ used} } keys %$hr;
> > }
> >
> > Anno
> > --
>
> Thanks Anno . I would rather go with this than with Gunnars suggestion
> since in my real program which is much larger, the %dbspaces hash is
> not a global variable and I would prefer to pass it to the sort
> function explicitly.
>
> I had kind of figured this might be the besty way to go but I still
> don't understand how $a and $b get passed through to the sort

The system takes care of $a and $b, they're none of your business.

> subroutine though and therefore cannot figure out how to pass
> aditional parameters.

What have $a and $b to do with additional parameters?

> I tried changing
>
> my @sort_by_used_keys = sort sort_by_used ( keys %dbspaces);
^^^^
What is the second sort() for? It will destroy the order sort_by_used()
has established.

> to
>
> my @sort_by_used_keys = sort sort_by_used ( \%dbspaces, keys
> %dbspaces);
>
> but then in the sort_by_used sub I don't appear to get the value ..
>
> sub sort_by_used
> {
> my $hr = shift;
>
> print Dumper $hr;
> return(0);
> }
>
> gives me
>
> $VAR1 = undef;

Not if you call sort_by_used as shown, it would show the content of
%dbspaces.

If you want to pass the keys as parameters, the sub should in some
way make use of them. Yours doesn't.

sub sort_by_used {
my $hr = shift;
sort { $hr->{ $a}->{ used} <=> $hr->{ $b}->{ used} @_;
}

Anno
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