Containers in Perl 6

Containers in Perl 6

In the third article in this series comparing Perl 5 to Perl 6, learn how to handle references, bindings, and containers in Perl 6.

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In the first article in this series comparing Perl 5 to Perl 6, we looked into some of the issues you might encounter when migrating code into Perl 6. In the second article, we examined how garbage collection works in Perl 6. Here, in the third article, we'll focus on Perl 5's references and how they're handled in Perl 6, and introduce the concepts of binding and containers.

References

There are no references in Perl 6, which is surprising to many people used to Perl 5's semantics. But worry not: because there are no references, you don't have to worry about whether something should be de-referenced or not.

# Perl 5
my $foo = \@bar;   # must add reference \ to make $foo a reference to @bar
say @bar[1];       # no dereference needed
say $foo->[1];     # must add dereference ->

# Perl 6
my $foo = @bar;    # $foo now contains @bar
say @bar[1];       # no dereference needed, note: sigil does not change
say $foo[1];       # no dereference needed either

One could argue that everything in Perl 6 is a reference. Coming from Perl 5 (where an object is a blessed reference), this would be a logical conclusion about Perl 6 where everything is an object (or can be considered one). But that wouldn't do justice to the situation in Perl 6 and would hinder you in understanding how things work in Perl 6. Beware of false friends!

Binding

Before we get to assignment, it is important to understand the concept of binding in Perl 6. You can bind something explicitly to something else using the := operator. When you define a lexical variable, you can bind a value to it:

my $foo := 42;  # note: := instead of =

Simply put, this creates a key with the name "$foo" in the lexical pad (lexpad) (which you could consider a compile-time hash that contains information about things that are visible in that lexical scope) and makes 42 its literal value. Because this is a literal constant, you can't change it. Trying to do so will cause an exception. So don't do that!

This binding operation is used under the hood in many situations, for instance when iterating:

my @a = 0..9;    # can also be written as ^10
say @a;          # [0 1 2 3 4 5 6 7 8 9]
for @a { $_++ }  # $_ is bound to each array element and incremented
say @a;          # [1 2 3 4 5 6 7 8 9 10]

If you try to iterate over a constant list, then $_ is bound to the literal values, which you can not increment:

for 0..9 { $_++ }  # error: requires mutable arguments

Assignment

If you compare "create a lexical variable and assign to it" in Perl 5 and Perl 6, it looks the same on the outside:

my $bar = 56;  # both Perl 5 and Perl 6

In Perl 6, this also creates a key with the name "$bar" in the lexpad. But instead of directly binding the value to that lexpad entry, a container (a Scalar object) is created for you and that is bound to the lexpad entry of "$bar". Then, 56 is stored as the value in that container. In pseudo-code, you can think of this as:

my $bar := Scalar.new( value => 56 );

Notice that the Scalar object is bound, not assigned. The closest thing to this in Perl 5 is a tied scalar. But of course "= 56" is much less to type!

Data structures such as Array and Hash also automatically put values in containers bound to the structure.

my @a;       # empty Array
@a[5] = 42;  # bind a Scalar container to 6th element and put 42 in it

Containers

The Scalar container object is invisible for most operations in Perl 6, so most of the time you don't have to think about it. For instance, whenever you call a subroutine (or a method) with a variable as an argument, it will bind to the value in the container. And because you cannot assign to a value, you get:

sub frobnicate($this) {
    $this = 42;
}
my $foo = 666;
frobnicate($foo); # Cannot assign to a readonly variable or a value

If you want to allow assigning to the outer value, you can add the is rw trait to the variable in the signature. This will bind the variable in the signature to the container of the variable specified, thus allowing assignment:

sub oknicate($this is rw) {
    $this = 42;
}
my $foo = 666;
oknicate($foo); # no problem
say $foo;       # 42

Proxy

Conceptually, the Scalar object in Perl 6 has a FETCH method (for producing the value in the object) and a STORE method (for changing the value in the object), just like a tied scalar in Perl 5.

Suppose you later assign the value 768 to the $bar variable:

$bar = 768;

What happens is conceptually the equivalent of:

$bar.STORE(768);

Suppose you want to add 20 to the value in $bar:

$bar = $bar + 20;

What happens conceptually is:

$bar.STORE( $bar.FETCH + 20 );

If you like to specify your own FETCH and STORE methods on a container, you can do that by binding to a Proxy object. For example, to create a variable that will always report twice the value that was assigned to it:

my $double := do {  # $double now a Proxy, rather than a Scalar container
    my $value;
    Proxy.new(
      FETCH => method ()     { $value + $value },
      STORE => method ($new) { $value = $new }
    )
}

Note that you will need an extra variable to keep the value stored in such a container.

Constraints and default

Apart from the value, a Scalar also contains extra information such as the type constraint and default value. Take this definition:

my Int $baz is default(42) = 666;

It creates a Scalar bound with the name "$baz" to the lexpad, constrains the values in that container to types that successfully smartmatch with Int, sets the default value of the container to 42, and puts the value 666 in the container.

Assigning a string to that variable will fail because of the type constraint:

$baz = "foo";
# Type check failed in assignment to $baz; expected Int but got Str ("foo")

If you do not give a type constraint when you define a variable, then the Any type will be assumed. If you do not specify a default value, then the type constraint will be assumed.

Assigning Nil (the Perl 6 equivalent of Perl 5's undef) to that variable will reset it to the default value:

say $baz;   # 666
$baz = Nil;
say $baz;   # 42

Summary

Perl 5 has values and references to values. Perl 6 has no references, but it has values and containers. There are two types of containers in Perl 6: Proxy (which is much like a tied scalar in Perl 5) and Scalar. Simply stated, a variable, as well as an element of a List, Array, or Hash, is either a value (if it is bound), or a container (if it is assigned). Whenever a subroutine (or method) is called, the given arguments are de-containerized and bound to the parameters of the subroutine (unless told to do otherwise). A container also keeps information such as type constraints and a default value. Assigning Nil to a variable will return it to its default value, which is Any if you do not specify a type constraint.

About the author

Elizabeth Mattijsen
Elizabeth Mattijsen - Elizabeth Mattijsen has been programming for a living since 1978 in various (mostly now defunct) programming languages before she started programming in Perl 4. In 1994 she started the first commercial web site development company in the Netherlands, using Perl 5 as the main programming language. From 2003 onwards, she was involved in the rapid growth of an online Hotel Reservation service. In 2012 she got more directly involved in the development of Rakudo Perl 6. From 2015 she has also...