-
=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,
+'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 it
-can be hard to tell what is important and what isn't.
+is quite complete, and sometimes people find that a problem, because
+it can be hard to tell what is important and what isn't.
Fortunately, you only need to know 10% of what's in the main page to get
90% of the benefit. This page will show you that 10%.
=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 state names, like this:
+have a file of city and country names, like this:
- Chicago, Illinois
- New York, New York
- Albany, New York
- Springfield, Illinois
- Trenton, New Jersey
- Evanston, Illinois
+ Chicago, USA
+ Frankfurt, Germany
+ Berlin, Germany
+ Washington, USA
+ Helsinki, Finland
+ New York, USA
-and you want to produce an output like this, with each state mentioned
-once, and then an alphabetical list of the cities in that state:
+and you want to produce an output like this, with each country mentioned
+once, and then an alphabetical list of the cities in that country:
- Illinois: Chicago, Evanston, Springfield.
- New Jersey: Trenton.
- New York: Albany, New York.
+ Finland: Helsinki.
+ Germany: Berlin, Frankfurt.
+ USA: Chicago, New York, Washington.
-The natural way to do this is to have a hash whose keys are state
-names. Associated with each state name key is a list of the cities in
-that state. Each time you read a line of input, split it into a state
+The natural way to do this is to have a hash whose keys are country
+names. Associated with each country name key is a list of the cities in
+that country. Each time you read a line of input, split it into a country
and a city, look up the list of cities already known to be in that
-state, and append the new city to the list. When you're done reading
+country, and append the new city to the list. When you're done reading
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
=head1 The Solution
-Unfortunately, by the time Perl 5 rolled around, we were already stuck
-with this design: Hash values must be scalars. The solution to this is
+By the time Perl 5 rolled around, we were already stuck with this
+design: Hash values must be scalars. The solution to this is
references.
A reference is a scalar value that I<refers to> an entire array or an
-entire hash (or to just about anything else.) Names are one kind of
-reference that you're already familiar with. Think of the President:
-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 "Bill Clinton".
+entire hash (or to just about anything else). Names are one kind of
+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 "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 "Bill Clinton", 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
references to arrays, and it'll act a lot like a hash of arrays, and
it'll be just as useful as a hash of arrays.
-We'll come back to this city-state problem later, after we've seen
+We'll come back to this city-country problem later, after we've seen
some syntax for managing references.
=head2 Making References
-B<Make Rule 1>
+=head3 B<Make Rule 1>
If you put a C<\> in front of a variable, you get a
reference to that variable.
$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:
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 };
# $href now holds a reference to a hash
The first line is an abbreviation for the following two lines, except
that it doesn't create the superfluous array variable C<@array>.
+If you write just C<[]>, you get a new, empty anonymous array.
+If you write just C<{}>, you get a new, empty anonymous hash.
+
=head2 Using References
value, and we've seen that you can store it as a scalar and get it back
again just like any scalar. There are just two more ways to use it:
-B<Use Rule 1>
+=head3 B<Use Rule 1>
-If C<$aref> contains a reference to an array, then you
-can put C<{$aref}> anywhere you would normally put the name of an
-array. For example, C<@{$aref}> instead of C<@array>.
+You can always use an array reference, in curly braces, in place of
+the name of an array. For example, C<@{$aref}> instead of C<@array>.
Here are some examples of that:
On each line are two expressions that do the same thing. The
-left-hand versions operate on the array C<@a>, and the right-hand
-versions operate on the array that is referred to by C<$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:
$h{'red'} ${$href}{'red'} An element of the hash
$h{'red'} = 17 ${$href}{'red'} = 17 Assigning an element
+Whatever you want to do with a reference, B<Use Rule 1> tells you how
+to do it. You just write the Perl code that you would have written
+for doing the same thing to a regular array or hash, and then replace
+the array or hash name with C<{$reference}>. "How do I loop over an
+array when all I have is a reference?" Well, to loop over an array, you
+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
+reference?" First write the code for printing out a hash:
+
+ for my $key (keys %hash) {
+ print "$key => $hash{$key}\n";
+ }
+
+And then replace the hash name with the reference:
+
+ for my $key (keys %{$href}) {
+ print "$key => ${$href}{$key}\n";
+ }
+
+=head3 B<Use Rule 2>
-B<Use Rule 2>
+B<Use Rule 1> is all you really need, because it tells you how to do
+absolutely everything you ever need to do with references. But the
+most common thing to do with an array or a hash is to extract a single
+element, and the B<Use Rule 1> notation is cumbersome. So there is an
+abbreviation.
-C<${$aref}[3]> is too hard to read, so you can write C<$aref-E<gt>[3]>
+C<${$aref}[3]> is too hard to read, so you can write C<< $aref->[3] >>
instead.
C<${$href}{red}> is too hard to read, so you can write
-C<$href-E<gt>{red}> instead.
+C<< $href->{red} >> instead.
-Most often, when you have an array or a hash, you want to get or set a
-single element from it. C<${$aref}[3]> and C<${$href}{'red'}> have
-too much punctuation, and Perl lets you abbreviate.
-
-If C<$aref> holds a reference to an array, then C<$aref-E<gt>[3]> is
+If C<$aref> holds a reference to an array, then C<< $aref->[3] >> is
the fourth element of the array. Don't confuse this with C<$aref[3]>,
which is the fourth element of a totally different array, one
deceptively named C<@aref>. C<$aref> and C<@aref> are unrelated the
same way that C<$item> and C<@item> are.
-Similarly, C<$href-E<gt>{'red'}> is part of the hash referred to by
+Similarly, C<< $href->{'red'} >> is part of the hash referred to by
the scalar variable C<$href>, perhaps even one with no name.
C<$href{'red'}> is part of the deceptively named C<%href> hash. It's
-easy to forget to leave out the C<-E<gt>>, and if you do, you'll get
+easy to forget to leave out the C<< -> >>, and if you do, you'll get
bizarre results when your program gets array and hash elements out of
totally unexpected hashes and arrays that weren't the ones you wanted
to use.
-=head1 An Example
+=head2 An Example
Let's see a quick example of how all this is useful.
C<$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<$a[1]-E<gt>[2]> to get the
-third element from that array. C<$a[1]-E<gt>[2]> is the 6.
-Similarly, C<$a[0]-E<gt>[1]> is the 2. What we have here is like a
-two-dimensional array; you can write C<$a[ROW]-E<gt>[COLUMN]> to get
+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<< $a[ROW]->[COLUMN] >> to get
or set the element in any row and any column of the array.
The notation still looks a little cumbersome, so there's one more
-abbreviation:
+abbreviation:
-=head1 Arrow Rule
+=head2 Arrow Rule
In between two B<subscripts>, the arrow is optional.
-Instead of C<$a[1]-E<gt>[2]>, we can write C<$a[1][2]>; it means the
-same thing. Instead of C<$a[0]-E<gt>[1]>, we can write C<$a[0][1]>;
-it means the same thing.
+Instead of C<< $a[1]->[2] >>, we can write C<$a[1][2]>; it means the
+same thing. Instead of C<< $a[0]->[1] = 23 >>, we can write
+C<$a[0][1] = 23>; it means the same thing.
Now it really looks like two-dimensional arrays!
three-dimensional arrays, they let us write C<$x[2][3][5]> instead of
the unreadable C<${${$x[2]}[3]}[5]>.
-
=head1 Solution
-Here's the answer to the problem I posed the the beginning of the
-article, of reformatting a file of city and state names.
-
- 1 while (<>) {
- 2 chomp;
- 3 my ($city, $state) = split /, /;
- 4 push @{$table{$state}}, $city;
- 5 }
- 6
- 7 foreach $state (sort keys %table) {
- 8 print "$state: ";
- 9 my @cities = @{$table{$state}};
- 10 print join ', ', sort @cities;
- 11 print ".\n";
- 12 }
-
-
-The program has two pieces: Lines 1--5 read the input and build a
-data structure, and lines 7--12 analyze the data and print out the
-report.
-
-In the first part, line 4 is the important one. We're going to have a
-hash, C<%table>, whose keys are state names, and whose values are
-(references to) arrays of city names. After acquiring a city and
-state name, the program looks up C<$table{$state}>, which holds (a
-reference to) the list of cities seen in that state so far. Line 4 is
-totally analogous to
+Here's the answer to the problem I posed earlier, of reformatting a
+file of city and country names.
+
+ 1 my %table;
+
+ 2 while (<>) {
+ 3 chomp;
+ 4 my ($city, $country) = split /, /;
+ 5 $table{$country} = [] unless exists $table{$country};
+ 6 push @{$table{$country}}, $city;
+ 7 }
+
+ 8 foreach $country (sort keys %table) {
+ 9 print "$country: ";
+ 10 my @cities = @{$table{$country}};
+ 11 print join ', ', sort @cities;
+ 12 print ".\n";
+ 13 }
+
+
+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
+structure will look like this:
+
+
+ %table
+ +-------+---+
+ | | | +-----------+--------+
+ |Germany| *---->| Frankfurt | Berlin |
+ | | | +-----------+--------+
+ +-------+---+
+ | | | +----------+
+ |Finland| *---->| Helsinki |
+ | | | +----------+
+ +-------+---+
+ | | | +---------+------------+----------+
+ | USA | *---->| Chicago | Washington | New York |
+ | | | +---------+------------+----------+
+ +-------+---+
+
+We'll look at output first. Supposing we already have this structure,
+how do we print it out?
+
+ 8 foreach $country (sort keys %table) {
+ 9 print "$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<$country> in the hash
+and gets the value, which is a reference to an array of cities in that country.
+B<Use Rule 1> says that
+we can recover the array by saying
+C<@{$table{$country}}>. Line 10 is just like
- push @array, $city;
+ @cities = @array;
except that the name C<array> has been replaced by the reference
-C<{$table{$state}}>. The C<push> adds a city name to the end of the
-referred-to array.
+C<{$table{$country}}>. The C<@> tells Perl to get the entire array.
+Having gotten the list of cities, we sort it, join it, and print it
+out as usual.
+
+Lines 2-7 are responsible for building the structure in the first
+place. Here they are again:
+
+ 2 while (<>) {
+ 3 chomp;
+ 4 my ($city, $country) = split /, /;
+ 5 $table{$country} = [] unless exists $table{$country};
+ 6 push @{$table{$country}}, $city;
+ 7 }
+
+Lines 2-4 acquire a city and country name. Line 5 looks to see if the
+country is already present as a key in the hash. If it's not, the
+program uses the C<[]> notation (B<Make Rule 2>) to manufacture a new,
+empty anonymous array of cities, and installs a reference to it into
+the hash under the appropriate key.
+
+Line 6 installs the city name into the appropriate array.
+C<$table{$country}> now holds a reference to the array of cities seen
+in that country so far. Line 6 is exactly like
-In the second part, line 9 is the important one. Again,
-C<$table{$state}> is (a reference to) the list of cities in the state, so
-we can recover the original list, and copy it into the array C<@cities>,
-by using C<@{$table{$state}}>. Line 9 is totally analogous to
-
- @cities = @array;
+ push @array, $city;
except that the name C<array> has been replaced by the reference
-C<{$table{$state}}>. The C<@> tells Perl to get the entire array.
-
-The rest of the program is just familiar uses of C<chomp>, C<split>, C<sort>,
-C<print>, and doesn't involve references at all.
+C<{$table{$country}}>. The C<push> adds a city name to the end of the
+referred-to array.
-There's one fine point I skipped. Suppose the program has just read
-the first line in its input that happens to mention the state of Ohio.
-Control is at line 4, C<$state> is C<'Ohio'>, and C<$city> is
-C<'Cleveland'>. Since this is the first city in Ohio,
-C<$table{$state}> is undefined---in fact there isn't an C<'Ohio'> key
-in C<%table> at all. What does line 4 do here?
+There's one fine point I skipped. Line 5 is unnecessary, and we can
+get rid of it.
- 4 push @{$table{$state}}, $city;
+ 2 while (<>) {
+ 3 chomp;
+ 4 my ($city, $country) = split /, /;
+ 5 #### $table{$country} = [] unless exists $table{$country};
+ 6 push @{$table{$country}}, $city;
+ 7 }
+If there's already an entry in C<%table> for the current C<$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>?
This is Perl, so it does the exact right thing. It sees that you want
-to push C<Cleveland> onto an array that doesn't exist, so it helpfully
-makes a new, empty, anonymous array for you, installs it in the table,
-and then pushes C<Cleveland> onto it. This is called `autovivification'.
-
+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
+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
+element longer to hold the new city name.
=head1 The Rest
=item *
-In B<USE RULE 1>, you can often omit the curly braces. For example,
+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 jsut starting out, you might want to adopt
-the habit of always including the curly braces.
+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;
+
+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.
+
+To copy the array, use
+
+ $aref2 = [@{$aref1}];
+
+This uses C<[...]> notation to create a new anonymous array, and
+C<$aref2> is assigned a reference to the new array. The new array is
+initialized with the contents of the array referred to by C<$aref1>.
+
+Similarly, to copy an anonymous hash, you can use
-=item *
+ $href2 = {%{$href1}};
-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 HASH for hash references and
-ARRAYfor array references.
+=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
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<soft reference> or I<symbolic
+reference>. The declaration C<use strict 'refs'> 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> volume 3, #2.
-Reprinted with permission.
+This article originally appeared in I<The Perl Journal>
+( http://www.tpj.com/ ) volume 3, #2. Reprinted with permission.
The original title was I<Understand References Today>.
+=head2 Distribution Conditions
-=cut
+Copyright 1998 The Perl Journal.
+
+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
+encouraged to use this code in your own programs for fun or for profit
+as you see fit. A simple comment in the code giving credit would be
+courteous but is not required.
+
+
+
+
+=cut
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