-
=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:
=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<< $aref->[3] >>
instead.
C<${$href}{red}> is too hard to read, so you can write
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<< $aref->[3] >> is
the fourth element of the array. Don't confuse this with C<$aref[3]>,
which is the fourth element of a totally different array, one
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]->[2] >> to get the
+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<< $a[1]->[2] >>, we can write C<$a[1][2]>; it means the
-same thing. Instead of C<< $a[0]->[1] >>, we can write C<$a[0][1]>;
-it means the same thing.
+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 earlier, of reformatting a
file of city and country names.
- 1 while (<>) {
- 2 chomp;
- 3 my ($city, $country) = split /, /;
- 4 push @{$table{$country}}, $city;
- 5 }
- 6
- 7 foreach $country (sort keys %table) {
- 8 print "$country: ";
- 9 my @cities = @{$table{$country}};
- 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 country names, and whose values are
-(references to) arrays of city names. After acquiring a city and
-country name, the program looks up C<$table{$country}>, which holds (a
-reference to) the list of cities seen in that country so far. Line 4 is
-totally analogous to
+ 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{$country}}>. The C<push> adds a city name to the end of the
-referred-to array.
+C<{$table{$country}}>. The C<@> tells Perl to get the entire array.
+Having gotten the list of cities, we sort it, join it, and print it
+out as usual.
-In the second part, line 9 is the important one. Again,
-C<$table{$country}> is (a reference to) the list of cities in the country, so
-we can recover the original list, and copy it into the array C<@cities>,
-by using C<@{$table{$country}}>. Line 9 is totally analogous to
+Lines 2-7 are responsible for building the structure in the first
+place. Here they are again:
- @cities = @array;
+ 2 while (<>) {
+ 3 chomp;
+ 4 my ($city, $country) = split /, /;
+ 5 $table{$country} = [] unless exists $table{$country};
+ 6 push @{$table{$country}}, $city;
+ 7 }
-except that the name C<array> has been replaced by the reference
-C<{$table{$country}}>. The C<@> tells Perl to get the entire array.
+Lines 2-4 acquire a city and country name. Line 5 looks to see if the
+country is already present as a key in the hash. If it's not, the
+program uses the C<[]> notation (B<Make Rule 2>) to manufacture a new,
+empty anonymous array of cities, and installs a reference to it into
+the hash under the appropriate key.
-The rest of the program is just familiar uses of C<chomp>, C<split>, C<sort>,
-C<print>, and doesn't involve references at all.
+Line 6 installs the city name into the appropriate array.
+C<$table{$country}> now holds a reference to the array of cities seen
+in that country so far. Line 6 is exactly like
-There's one fine point I skipped. Suppose the program has just read
-the first line in its input that happens to mention Greece.
-Control is at line 4, C<$country> is C<'Greece'>, and C<$city> is
-C<'Athens'>. Since this is the first city in Greece,
-C<$table{$country}> is undefined---in fact there isn't an C<'Greece'> key
-in C<%table> at all. What does line 4 do here?
+ push @array, $city;
+
+except that the name C<array> has been replaced by the reference
+C<{$table{$country}}>. The C<push> adds a city name to the end of the
+referred-to array.
- 4 push @{$table{$country}}, $city;
+There's one fine point I skipped. Line 5 is unnecessary, and we can
+get rid of it.
+ 2 while (<>) {
+ 3 chomp;
+ 4 my ($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<Athens> 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<Athens> onto it. This is called `autovivification'.
-
+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 omit the curly brackets whenever the thing
+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.
-=item *
+=item *
+
+This doesn't copy the underlying array:
+
+ $aref2 = $aref1;
-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
-ARRAY for array references.
+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.
-=item *
+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
+
+ $href2 = {%{$href1}};
+
+=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 *
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>
-( 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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