PERL Beginners - passing hash data from one subroutine to another using refs

Author

passing hash data from one subroutine to another using refs

Brent Clark

2005-07-28, 9:07 am

Hi

Would someone please clear up / help me understand this.

I have a hash

my %allTariffData = {};

and in my sub

I have
my %tariffData = {};
do the processing etc
and then
return \%tariffData;

I now would like to use that data that in tariffData in another subroutine.

Would anyone be so kind as to share how I would go about this.

I can pass the whole hash, but that would not be right.

Kind Regards and thanks in advance
Brent Clark

Ankur Gupta

2005-07-28, 9:07 am

Brent Clark <mailto:bclark@eccotours.dyndns.org> wrote:
> Hi

Hi,
=20
> Would someone please clear up / help me understand this.

Sure.
=20
> I have a hash
>=20
> my %allTariffData =3D {};

I think you meant my %allTariffData =3D ();

> and in my sub
>=20
> I have
> my %tariffData =3D {};

Same here.. my %tariffData =3D ();

> do the processing etc
> and then
> return \%tariffData;

A reference of %tariffData is returned to the caller.
=20
> I now would like to use that data that in tariffData in another
> subroutine.=20

I guess your initial subroutine would be like this.

$ref_tariffData =3D func();
subroutine func returns refernce to a hash.

If you want to use it in another subroutine. simply pass the reference
of that hash to the subroutine.
func2($ref_tariffData);

sub func2 {
my ($ref_hash) =3D @_;
foreach my $key ( keys %$ref_hash ){
print "$key: $ref_hash->{$key}\n";
}
}

> Would anyone be so kind as to share how I would go about this.
>=20
> I can pass the whole hash, but that would not be right.

You can pass the whole hash but only if you are passing one hash.=20
return %hash; # OK
return %hash1, %hash2; # May not be OK

Please read the following perldocs:

perldoc perldsc
perldoc perlref
perldoc perllol
perldoc perldata

It will make the things a lot clearer to you.

--Ankur

I don't suffer from insanity, I enjoy every minute of it!

Tom Allison

2005-07-29, 10:00 pm

> sub func2 {
> my ($ref_hash) = @_;
> foreach my $key ( keys %$ref_hash ){
> print "$key: $ref_hash->{$key}\n";
> }
> }
>
>

What you can't do here is:

sub func2 {
my %hash = @_;
....
}

You've broken the reference back to the original hash.
you will modify the elements in %hash, but not in your %tariffData

But you can declare a reference

my %tarrifData;
func1(\%tarrifData);
# %tarrifData has been changed
func2(\%tarrifData);
func3(\%tarrifData); # does not modify %tarrifData

sub func1 {
my $ref = shift; # points to %tarrifData
# Do Something
}

sub func2 {
my $ref = shift; # points again to %tarrifData
# Do something else
}

sub func3 {
my %hash = %{$_[0]}; # no longer points to %tarrifData
undef %hash; # who cares?
}

now go read the perldocs.. :)

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