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
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>.