Does Perl6 support dependent types?

Question

I was recently looking at the wikipedia page for dependent types, and I was wondering; does Perl 6 actually introduce dependent types? I can\'t seem to find a reliable source cl

Answer 1

Arguably yes as subsets are types that may depend on arbitrary conditions. However, the type system would be classified as unsound as type invariants are not enforced.

In particular, a variable's type constraint is only checked on assignment, so modifications to an object that make it drop from a subset will lead to a variable holding an object it should not be able to, eg

subset OrderedList of List where [<=] @$_;

my OrderedList $list = [1, 2, 3];
$list[0] = 42;
say $list ~~ OrderedList;

You can use some meta-object wizardry to make the object system automatically check the type after any method call by boxing objects in transparent guard objects.

A naive implementation could look like this:

class GuardHOW {
    has $.obj;
    has $.guard;
    has %!cache =
        gist => sub (Mu \this) {
            this.DEFINITE
                ?? $!obj.gist
                !! "({ self.name(this) })";
        },
        UNBOX => sub (Mu $) { $!obj };

    method find_method(Mu $, $name) {
        %!cache{$name} //= sub (Mu $, |args) {
            POST $!obj ~~ $!guard;
            $!obj."$name"(|args);
        }
    }

    method name(Mu $) { "Guard[{ $!obj.^name }]" }
    method type_check(Mu $, $type) { $!obj ~~ $type }
}

sub guard($obj, $guard) {
    use nqp;
    PRE $obj ~~ $guard;
    nqp::create(nqp::newtype(GuardHOW.new(:$obj, :$guard), 'P6int'));
}

This will make the following fail:

my $guarded-list = guard([1, 2, 3], OrderedList);
$guarded-list[0] = 42;