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perlref - Perl references and nested data structures


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.

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;

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".)

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

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.

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');

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.

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;

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.

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.

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.


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.