PERLREF(1)PERLREF(1)NAMEperlref - Perl references and nested data structures
DESCRIPTION
In Perl 4 it was difficult to represent complex data structures,
because all references had to be symbolic, and even that was difficult
to do when you wanted to refer to a variable rather than a symbol table
entry. Perl 5 not only makes it easier to use symbolic references to
variables, but lets you have "hard" references to any piece of data.
Any scalar may hold a hard reference. Since arrays and hashes contain
scalars, you can now easily build arrays of arrays, arrays of hashes,
hashes of arrays, arrays of hashes of functions, and so on.
Hard references are smart--they keep track of reference counts for you,
automatically freeing the thing referred to when its reference count
goes to zero. If that thing happens to be an object, the object is
destructed. See the perlobj manpage for more about objects. (In a
sense, everything in Perl is an object, but we usually reserve the word
for references to objects that have been officially "blessed" into a
class package.)
A symbolic reference contains the name of a variable, just as a
symbolic link in the filesystem merely contains the name of a file.
The *glob notation is a kind of symbolic reference. Hard references
are more like hard links in the file system: merely another way at
getting at the same underlying object, irrespective of its name.
"Hard" references are easy to use in Perl. There is just one
overriding principle: Perl does no implicit referencing or
dereferencing. When a scalar is holding a reference, it always behaves
as a scalar. It doesn't magically start being an array or a hash
unless you tell it so explicitly by dereferencing it.
References can be constructed several ways.
1. By using the backslash operator on a variable, subroutine, or
value. (This works much like the & (address-of) operator works in
C.) Note that this typically creates ANOTHER reference to a
variable, since there's already a reference to the variable in the
symbol table. But the symbol table reference might go away, and
you'll still have the reference that the backslash returned. Here
are some examples:
$scalarref = \$foo;
$arrayref = \@ARGV;
$hashref = \%ENV;
$coderef = \&handler;
2. A reference to an anonymous array can be constructed using square
brackets:
$arrayref = [1, 2, ['a', 'b', 'c']];
Here we've constructed a reference to an anonymous array of three
elements whose final element is itself reference to another
anonymous array of three elements. (The multidimensional syntax
described later can be used to access this. For example, after the
above, $arrayref->[2][1] would have the value "b".)
3. A reference to an anonymous hash can be constructed using curly
brackets:
$hashref = {
'Adam' => 'Eve',
'Clyde' => 'Bonnie',
};
Anonymous hash and array constructors can be intermixed freely to
produce as complicated a structure as you want. The
multidimensional syntax described below works for these too. The
values above are literals, but variables and expressions would work
just as well, because assignment operators in Perl (even within
local() or my()) are executable statements, not compile-time
declarations.
Because curly brackets (braces) are used for several other things
including BLOCKs, you may occasionally have to disambiguate braces
at the beginning of a statement by putting a + or a return in front
so that Perl realizes the opening brace isn't starting a BLOCK.
The economy and mnemonic value of using curlies is deemed worth
this occasional extra hassle.
For example, if you wanted a function to make a new hash and return
a reference to it, you have these options:
sub hashem { { @_ } } # silently wrong
sub hashem { +{ @_ } } # ok
sub hashem { return { @_ } } # ok
4. A reference to an anonymous subroutine can be constructed by using
sub without a subname:
$coderef = sub { print "Boink!\n" };
Note the presence of the semicolon. Except for the fact that the
code inside isn't executed immediately, a sub {} is not so much a
declaration as it is an operator, like do{} or eval{}. (However,
no matter how many times you execute that line (unless you're in an
eval("...")), $coderef will still have a reference to the SAME
anonymous subroutine.)
Anonymous subroutines act as closures with respect to my()
variables, that is, variables visible lexically within the current
scope. Closure is a notion out of the Lisp world that says if you
define an anonymous function in a particular lexical context, it
pretends to run in that context even when it's called outside of
the context.
In human terms, it's a funny way of passing arguments to a
subroutine when you define it as well as when you call it. It's
useful for setting up little bits of code to run later, such as
callbacks. You can even do object-oriented stuff with it, though
Perl provides a different mechanism to do that already--see the
perlobj manpage.
You can also think of closure as a way to write a subroutine
template without using eval. (In fact, in version 5.000, eval was
the only way to get closures. You may wish to use "require 5.001"
if you use closures.)
Here's a small example of how closures works:
sub newprint {
my $x = shift;
return sub { my $y = shift; print "$x, $y!\n"; };
}
$h = newprint("Howdy");
$g = newprint("Greetings");
# Time passes...
&$h("world");
&$g("earthlings");
This prints
Howdy, world!
Greetings, earthlings!
Note particularly that $x continues to refer to the value passed
into newprint() *despite* the fact that the "my $x" has seemingly
gone out of scope by the time the anonymous subroutine runs.
That's what closure is all about.
This only applies to lexical variables, by the way. Dynamic
variables continue to work as they have always worked. Closure is
not something that most Perl programmers need trouble themselves
about to begin with.
5. References are often returned by special subroutines called
constructors. Perl objects are just references to a special kind
of object that happens to know which package it's associated with.
Constructors are just special subroutines that know how to create
that association. They do so by starting with an ordinary
reference, and it remains an ordinary reference even while it's
also being an object. Constructors are customarily named new(),
but don't have to be:
$objref = new Doggie (Tail => 'short', Ears => 'long');
6. References of the appropriate type can spring into existence if you
dereference them in a context that assumes they exist. Since we
haven't talked about dereferencing yet, we can't show you any
examples yet.
That's it for creating references. By now you're probably dying to
know how to use references to get back to your long-lost data. There
are several basic methods.
1. Anywhere you'd put an identifier as part of a variable or
subroutine name, you can replace the identifier with a simple
scalar variable containing a reference of the correct type:
$bar = $$scalarref;
push(@$arrayref, $filename);
$$arrayref[0] = "January";
$$hashref{"KEY"} = "VALUE";
&$coderef(1,2,3);
It's important to understand that we are specifically NOT
dereferencing $arrayref[0] or $hashref{"KEY"} there. The
dereference of the scalar variable happens BEFORE it does any key
lookups. Anything more complicated than a simple scalar variable
must use methods 2 or 3 below. However, a "simple scalar" includes
an identifier that itself uses method 1 recursively. Therefore,
the following prints "howdy".
$refrefref = \\\"howdy";
print $$$$refrefref;
2. Anywhere you'd put an identifier as part of a variable or
subroutine name, you can replace the identifier with a BLOCK
returning a reference of the correct type. In other words, the
previous examples could be written like this:
$bar = ${$scalarref};
push(@{$arrayref}, $filename);
${$arrayref}[0] = "January";
${$hashref}{"KEY"} = "VALUE";
&{$coderef}(1,2,3);
Admittedly, it's a little silly to use the curlies in this case,
but the BLOCK can contain any arbitrary expression, in particular,
subscripted expressions:
&{ $dispatch{$index} }(1,2,3); # call correct routine
Because of being able to omit the curlies for the simple case of
$$x, people often make the mistake of viewing the dereferencing
symbols as proper operators, and wonder about their precedence. If
they were, though, you could use parens instead of braces. That's
not the case. Consider the difference below; case 0 is a short-
hand version of case 1, NOT case 2:
$$hashref{"KEY"} = "VALUE"; # CASE 0
${$hashref}{"KEY"} = "VALUE"; # CASE 1
${$hashref{"KEY"}} = "VALUE"; # CASE 2
${$hashref->{"KEY"}} = "VALUE"; # CASE 3
Case 2 is also deceptive in that you're accessing a variable called
%hashref, not dereferencing through $hashref to the hash it's
presumably referencing. That would be case 3.
3. The case of individual array elements arises often enough that it
gets cumbersome to use method 2. As a form of syntactic sugar, the
two lines like that above can be written:
$arrayref->[0] = "January";
$hashref->{"KEY"} = "VALUE";
The left side of the array can be any expression returning a
reference, including a previous dereference. Note that $array[$x]
is NOT the same thing as $array->[$x] here:
$array[$x]->{"foo"}->[0] = "January";
This is one of the cases we mentioned earlier in which references
could spring into existence when in an lvalue context. Before this
statement, $array[$x] may have been undefined. If so, it's
automatically defined with a hash reference so that we can look up
{"foo"} in it. Likewise $array[$x]->{"foo"} will automatically get
defined with an array reference so that we can look up [0] in it.
One more thing here. The arrow is optional BETWEEN brackets
subscripts, so you can shrink the above down to
$array[$x]{"foo"}[0] = "January";
Which, in the degenerate case of using only ordinary arrays, gives
you multidimensional arrays just like C's:
$score[$x][$y][$z] += 42;
Well, okay, not entirely like C's arrays, actually. C doesn't know
how to grow its arrays on demand. Perl does.
4. If a reference happens to be a reference to an object, then there
are probably methods to access the things referred to, and you
should probably stick to those methods unless you're in the class
package that defines the object's methods. In other words, be
nice, and don't violate the object's encapsulation without a very
good reason. Perl does not enforce encapsulation. We are not
totalitarians here. We do expect some basic civility though.
The ref() operator may be used to determine what type of thing the
reference is pointing to. See the perlfunc manpage.
The bless() operator may be used to associate a reference with a
package functioning as an object class. See the perlobj manpage.
A type glob may be dereferenced the same way a reference can, since the
dereference syntax always indicates the kind of reference desired. So
${*foo} and ${\$foo} both indicate the same scalar variable.
Here's a trick for interpolating a subroutine call into a string:
print "My sub returned ${\mysub(1,2,3)}\n";
The way it works is that when the ${...} is seen in the double-quoted
string, it's evaluated as a block. The block executes the call to
mysub(1,2,3), and then takes a reference to that. So the whole block
returns a reference to a scalar, which is then dereferenced by ${...}
and stuck into the double-quoted string.
Symbolic references
We said that references spring into existence as necessary if they are
undefined, but we didn't say what happens if a value used as a
reference is already defined, but ISN'T a hard reference. If you use
it as a reference in this case, it'll be treated as a symbolic
reference. That is, the value of the scalar is taken to be the NAME of
a variable, rather than a direct link to a (possibly) anonymous value.
People frequently expect it to work like this. So it does.
$name = "foo";
$$name = 1; # Sets $foo
${$name} = 2; # Sets $foo
${$name x 2} = 3; # Sets $foofoo
$name->[0] = 4; # Sets $foo[0]
@$name = (); # Clears @foo
&$name(); # Calls &foo() (as in Perl 4)
$pack = "THAT";
${"${pack}::$name"} = 5; # Sets $THAT::foo without eval
This is very powerful, and slightly dangerous, in that it's possible to
intend (with the utmost sincerity) to use a hard reference, and
accidentally use a symbolic reference instead. To protect against
that, you can say
use strict 'refs';
and then only hard references will be allowed for the rest of the
enclosing block. An inner block may countermand that with
no strict 'refs';
Only package variables are visible to symbolic references. Lexical
variables (declared with my()) aren't in a symbol table, and thus are
invisible to this mechanism. For example:
local($value) = 10;
$ref = \$value;
{
my $value = 20;
print $$ref;
}
This will still print 10, not 20. Remember that local() affects
package variables, which are all "global" to the package.
Not-so-symbolic references
A new feature contributing to readability in 5.001 is that the brackets
around a symbolic reference behave more like quotes, just as they
always have within a string. That is,
$push = "pop on ";
print "${push}over";
has always meant to print "pop on over", despite the fact that push is
a reserved word. This has been generalized to work the same outside of
quotes, so that
print ${push} . "over";
and even
print ${ push } . "over";
will have the same effect. (This would have been a syntax error in
5.000, though Perl 4 allowed it in the spaceless form.) Note that this
construct is not considered to be a symbolic reference when you're
using strict refs:
use strict 'refs';
${ bareword }; # Okay, means $bareword.
${ "bareword" }; # Error, symbolic reference.
Similarly, because of all the subscripting that is done using single
words, we've applied the same rule to any bareword that is used for
subscripting a hash. So now, instead of writing
$array{ "aaa" }{ "bbb" }{ "ccc" }
you can just write
$array{ aaa }{ bbb }{ ccc }
and not worry about whether the subscripts are reserved words. In the
rare event that you do wish to do something like
$array{ shift }
you can force interpretation as a reserved word by adding anything that
makes it more than a bareword:
$array{ shift() }
$array{ +shift }
$array{ shift @_ }
The -w switch will warn you if it interprets a reserved word as a
string. But it will no longer warn you about using lowercase words,
since the string is effectively quoted.
WARNING
You may not (usefully) use a reference as the key to a hash. It will
be converted into a string:
$x{ \$a } = $a;
If you try to dereference the key, it won't do a hard dereference, and
you won't accomplish what you're attemping.
Further Reading
Besides the obvious documents, source code can be instructive. Some
rather pathological examples of the use of references can be found in
the t/op/ref.t regression test in the Perl source directory.
3rd Berkeley DistributionPERLREF(1)
