-
=head1 NAME
perlreftut - Mark's very short tutorial about references
One problem that came up all the time in Perl 4 was how to represent a
hash whose values were lists. Perl 4 had hashes, of course, but the
-values had to be scalars; they couldn't be lists.
+values had to be scalars; they couldn't be lists.
Why would you want a hash of lists? Let's take a simple example: You
have a file of city and country names, like this:
$aref = \@array; # $aref now holds a reference to @array
$href = \%hash; # $href now holds a reference to %hash
+ $sref = \$scalar; # $sref now holds a reference to $scalar
Once the reference is stored in a variable like $aref or $href, you
can copy it or store it just the same as any other scalar value:
that array. C<{ ITEMS }> makes a new, anonymous hash, and returns a
reference to that hash.
- $aref = [ 1, "foo", undef, 13 ];
+ $aref = [ 1, "foo", undef, 13 ];
# $aref now holds a reference to an array
- $href = { APR => 4, AUG => 8 };
+ $href = { APR => 4, AUG => 8 };
# $href now holds a reference to a hash
=head3 B<Use Rule 2>
-B<Use Rule 1> is all you really need, because it tells you how to to
+B<Use Rule 1> is all you really need, because it tells you how to do
absolutely everything you ever need to do with references. But the
most common thing to do with an array or a hash is to extract a single
element, and the B<Use Rule 1> notation is cumbersome. So there is an
or set the element in any row and any column of the array.
The notation still looks a little cumbersome, so there's one more
-abbreviation:
+abbreviation:
=head2 Arrow Rule
%table
- +-------+---+
+ +-------+---+
| | | +-----------+--------+
|Germany| *---->| Frankfurt | Berlin |
| | | +-----------+--------+
referred-to array.
There's one fine point I skipped. Line 5 is unnecessary, and we can
-get rid of it.
+get rid of it.
2 while (<>) {
3 chomp;
This doesn't copy the underlying array:
- $aref2 = $aref1;
+ $aref2 = $aref1;
-You get two references to the same array. If you modify
+You get two references to the same array. If you modify
C<< $aref1->[23] >> and then look at
-C<< $aref2->[23] >> you'll see the change.
+C<< $aref2->[23] >> you'll see the change.
To copy the array, use
$href2 = {%{$href1}};
-=item *
+=item *
To see if a variable contains a reference, use the C<ref> function. It
returns true if its argument is a reference. Actually it's a little
better than that: It returns C<HASH> for hash references and C<ARRAY>
for array references.
-=item *
+=item *
If you try to use a reference like a string, you get strings like
Author: Mark Jason Dominus, Plover Systems (C<mjd-perl-ref+@plover.com>)
This article originally appeared in I<The Perl Journal>
-( http://www.tpj.com/ ) volume 3, #2. Reprinted with permission.
+( http://www.tpj.com/ ) volume 3, #2. Reprinted with permission.
The original title was I<Understand References Today>.