-
=head1 NAME
perlreftut - Mark's very short tutorial about references
One of the most important new features in Perl 5 was the capability to
manage complicated data structures like multidimensional arrays and
nested hashes. To enable these, Perl 5 introduced a feature called
-`references', and using references is the key to managing complicated,
+I<references>, and using references is the key to managing complicated,
structured data in Perl. Unfortunately, there's a lot of funny syntax
to learn, and the main manual page can be hard to follow. The manual
is quite complete, and sometimes people find that a problem, because
=head1 Who Needs Complicated Data Structures?
-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.
+One problem that comes up all the time is needing a hash whose values are
+lists. Perl has hashes, of course, but the values have to be scalars;
+they can'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:
the input, iterate over the hash as usual, sorting each list of cities
before you print it out.
-If hash values can't be lists, you lose. In Perl 4, hash values can't
-be lists; they can only be strings. You lose. You'd probably have to
+If hash values couldn't be lists, you lose. You'd probably have to
combine all the cities into a single string somehow, and then when
time came to write the output, you'd have to break the string into a
list, sort the list, and turn it back into a string. This is messy
reference that you're already familiar with. Think of the President
of the United States: a messy, inconvenient bag of blood and bones.
But to talk about him, or to represent him in a computer program, all
-you need is the easy, convenient scalar string "George Bush".
+you need is the easy, convenient scalar string "Barack Obama".
References in Perl are like names for arrays and hashes. They're
Perl's private, internal names, so you can be sure they're
-unambiguous. Unlike "George Bush", a reference only refers to one
+unambiguous. Unlike "Barack Obama", a reference only refers to one
thing, and you always know what it refers to. If you have a reference
to an array, you can recover the entire array from it. If you have a
reference to a hash, you can recover the entire hash. But the
$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:
string C<"\n"> or the number 80 without having to store it in a named
variable first.
-B<Make Rule 2>
+=head3 B<Make Rule 2>
C<[ ITEMS ]> makes a new, anonymous array, and returns a reference to
-that array. C<{ ITEMS }> makes a new, anonymous hash. and returns a
+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 };
# $href now holds a reference to a hash
On each line are two expressions that do the same thing. The
-left-hand versions operate on the array C<@a>, and the right-hand
-versions operate on the array that is referred to by C<$aref>, but
-once they find the array they're operating on, they do the same things
-to the arrays.
+left-hand versions operate on the array C<@a>. The right-hand
+versions operate on the array that is referred to by C<$aref>. Once
+they find the array they're operating on, both versions do the same
+things to the arrays.
Using a hash reference is I<exactly> the same:
would write
for my $element (@array) {
- ...
+ ...
}
so replace the array name, C<@array>, with the reference:
for my $element (@{$aref}) {
- ...
+ ...
}
"How do I print out the contents of a hash when all I have is a
=head3 B<Use Rule 2>
-B<Use Rule 1> is all you really need, because it tells you how to to
-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
-abbreviation.
+L<B<Use Rule 1>|/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 L<B<Use Rule 1>|/B<Use Rule 1>> notation is
+cumbersome. So there is an abbreviation.
C<${$aref}[3]> is too hard to read, so you can write C<< $aref->[3] >>
instead.
[7, 8, 9]
);
-@a is an array with three elements, and each one is a reference to
+C<@a> is an array with three elements, and each one is a reference to
another array.
C<$a[1]> is one of these references. It refers to an array, the array
containing C<(4, 5, 6)>, and because it is a reference to an array,
-B<Use Rule 2> says that we can write C<< $a[1]->[2] >> to get the
-third element from that array. C<< $a[1]->[2] >> is the 6.
+L<B<Use Rule 2>|/B<Use Rule 2>> says that we can write C<< $a[1]->[2] >>
+to get the third element from that array. C<< $a[1]->[2] >> is the 6.
Similarly, C<< $a[0]->[1] >> is the 2. What we have here is like a
-two-dimensional array; you can write C<< $a[ROW]->[COLUMN] >> to get
-or set the element in any row and any column of the array.
+two-dimensional array; you can write C<< $a[ROW]->[COLUMN] >> to get or
+set the element in any row and any column of the array.
The notation still looks a little cumbersome, so there's one more
-abbreviation:
+abbreviation:
=head2 Arrow Rule
1 my %table;
2 while (<>) {
- 3 chomp;
+ 3 chomp;
4 my ($city, $country) = split /, /;
5 $table{$country} = [] unless exists $table{$country};
6 push @{$table{$country}}, $city;
7 }
- 8 foreach $country (sort keys %table) {
+ 8 for my $country (sort keys %table) {
9 print "$country: ";
10 my @cities = @{$table{$country}};
11 print join ', ', sort @cities;
13 }
-The program has two pieces: Lines 2--7 read the input and build a data
+The program has two pieces: Lines 2-7 read the input and build a data
structure, and lines 8-13 analyze the data and print out the report.
We're going to have a hash, C<%table>, whose keys are country names,
and whose values are references to arrays of city names. The data
%table
- +-------+---+
+ +-------+---+
| | | +-----------+--------+
|Germany| *---->| Frankfurt | Berlin |
| | | +-----------+--------+
We'll look at output first. Supposing we already have this structure,
how do we print it out?
-C<%table> is an
-ordinary hash, and we get a list of keys from it, sort the keys, and
-loop over the keys as usual. The only use of references is in line 10.
-C<$table{$country}> looks up the key C<$country> in the hash
-and gets the value, which is a reference to an array of cities in that country.
-B<Use Rule 1> says that
-we can recover the array by saying
-C<@{$table{$country}}>. Line 10 is just like
+ 8 for my $country (sort keys %table) {
+ 9 print "$country: ";
+ 10 my @cities = @{$table{$country}};
+ 11 print join ', ', sort @cities;
+ 12 print ".\n";
+ 13 }
+
+C<%table> is an ordinary hash, and we get a list of keys from it, sort
+the keys, and loop over the keys as usual. The only use of references
+is in line 10. C<$table{$country}> looks up the key C<$country> in the
+hash and gets the value, which is a reference to an array of cities in
+that country. L<B<Use Rule 1>|/B<Use Rule 1>> says that we can recover
+the array by saying C<@{$table{$country}}>. Line 10 is just like
@cities = @array;
out as usual.
Lines 2-7 are responsible for building the structure in the first
-place; here they are again:
+place. Here they are again:
2 while (<>) {
- 3 chomp;
+ 3 chomp;
4 my ($city, $country) = split /, /;
5 $table{$country} = [] unless exists $table{$country};
6 push @{$table{$country}}, $city;
Lines 2-4 acquire a city and country name. Line 5 looks to see if the
country is already present as a key in the hash. If it's not, the
-program uses the C<[]> notation (B<Make Rule 2>) to manufacture a new,
-empty anonymous array of cities, and installs a reference to it into
-the hash under the appropriate key.
+program uses the C<[]> notation (L<B<Make Rule 2>|/B<Make Rule 2>>) to
+manufacture a new, empty anonymous array of cities, and installs a
+reference to it into the hash under the appropriate key.
Line 6 installs the city name into the appropriate array.
C<$table{$country}> now holds a reference to the array of cities seen
push @array, $city;
except that the name C<array> has been replaced by the reference
-C<{$table{$country}}>. The C<push> adds a city name to the end of the
-referred-to array.
+C<{$table{$country}}>. The L<C<push>|perlfunc/push ARRAY,LIST> adds a
+city name to the end of the referred-to array.
There's one fine point I skipped. Line 5 is unnecessary, and we can
-get rid of it.
+get rid of it.
2 while (<>) {
- 3 chomp;
+ 3 chomp;
4 my ($city, $country) = split /, /;
5 #### $table{$country} = [] unless exists $table{$country};
6 push @{$table{$country}}, $city;
If there's already an entry in C<%table> for the current C<$country>,
then nothing is different. Line 6 will locate the value in
-C<$table{$country}>, which is a reference to an array, and push
-C<$city> into the array. But
-what does it do when
-C<$country> holds a key, say C<Greece>, that is not yet in C<%table>?
+C<$table{$country}>, which is a reference to an array, and push C<$city>
+into the array. But what does it do when C<$country> holds a key, say
+C<Greece>, that is not yet in C<%table>?
This is Perl, so it does the exact right thing. It sees that you want
to push C<Athens> onto an array that doesn't exist, so it helpfully
makes a new, empty, anonymous array for you, installs it into
C<%table>, and then pushes C<Athens> onto it. This is called
-`autovivification'--bringing things to life automatically. Perl saw
-that they key wasn't in the hash, so it created a new hash entry
+I<autovivification>--bringing things to life automatically. Perl saw
+that the key wasn't in the hash, so it created a new hash entry
automatically. Perl saw that you wanted to use the hash value as an
array, so it created a new empty array and installed a reference to it
in the hash automatically. And as usual, Perl made the array one
=item *
-In B<USE RULE 1>, you can omit the curly brackets whenever the thing
-inside them is an atomic scalar variable like C<$aref>. For example,
-C<@$aref> is the same as C<@{$aref}>, and C<$$aref[1]> is the same as
-C<${$aref}[1]>. If you're just starting out, you may want to adopt
-the habit of always including the curly brackets.
+In L<B<Use Rule 1>|/B<Use Rule 1>>, you can omit the curly brackets
+whenever the thing inside them is an atomic scalar variable like
+C<$aref>. For example, C<@$aref> is the same as C<@{$aref}>, and
+C<$$aref[1]> is the same as C<${$aref}[1]>. If you're just starting
+out, you may want to adopt the habit of always including the curly
+brackets.
=item *
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
Similarly, to copy an anonymous hash, you can use
- $href = {%{$href}};
+ $href2 = {%{$href1}};
-=item *
+=item *
-To see if a variable contains a reference, use the `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.
+To see if a variable contains a reference, use the
+L<C<ref>|perlfunc/ref EXPR> 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
If you ever see a string that looks like this, you'll know you
printed out a reference by mistake.
-A side effect of this representation is that you can use C<eq> to see
-if two references refer to the same thing. (But you should usually use
-C<==> instead because it's much faster.)
+A side effect of this representation is that you can use
+L<C<eq>|perlop/Equality Operators> to see if two references refer to the
+same thing. (But you should usually use
+L<C<==>|perlop/Equality Operators> instead because it's much faster.)
=item *
You can use a string as if it were a reference. If you use the string
C<"foo"> as an array reference, it's taken to be a reference to the
-array C<@foo>. This is called a I<soft reference> or I<symbolic reference>.
+array C<@foo>. This is called a I<symbolic reference>. The declaration
+L<C<use strict 'refs'>|strict> disables this feature, which can cause
+all sorts of trouble if you use it by accident.
=back
=head1 Credits
-Author: Mark-Jason Dominus, Plover Systems (C<mjd-perl-ref+@plover.com>)
+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>.
Copyright 1998 The Perl Journal.
-When included as part of the Standard Version of Perl, or as part of
-its complete documentation whether printed or otherwise, this work may
-be distributed only under the terms of Perl's Artistic License. Any
-distribution of this file or derivatives thereof outside of that
-package require that special arrangements be made with copyright
-holder.
+This documentation is free; you can redistribute it and/or modify it
+under the same terms as Perl itself.
Irrespective of its distribution, all code examples in these files are
hereby placed into the public domain. You are permitted and
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