perl - 如何将取决于另一个属性的属性约束从触发器提升到类型细化？

Question

这有效：

use Moops;
class Foo :ro {
    use Types::Common::Numeric qw(PositiveOrZeroInt);
    has from => required => true, isa => PositiveOrZeroInt;
    has to => required => true, isa => PositiveOrZeroInt, trigger => method($to) {
        die 'must be from ≤ to' unless $self->from <= $to
    };
}
Foo->new(from => 0, to => 1); # ok
Foo->new(from => 1, to => 0); # "must be from ≤ to at …&quot;

我希望能够以某种方式使约束成为类型的一部分。

use Moops;
class Bar :ro {
    use Types::Common::Numeric qw(PositiveOrZeroInt);
    has from => required => true, isa => PositiveOrZeroInt;
    has to => required => true, isa => PositiveOrZeroInt->where(sub {
        $self->from <= $_
    });
}
Bar->new(from => 0, to => 1);
# Global symbol "$self" requires explicit package name
# (did you forget to declare "my $self"?)

我检查了wheresub只接收一个参数。

Answer 1

如果您想在类型检查中执行此操作，您可以将这两个属性组合成一个属性，该属性将是一个包含两个数字的 arrayref。

use Moops;

class Foo :ro {
    use Types::Standard qw(Tuple);
    use Types::Common::Numeric qw(PositiveOrZeroInt);
    
    has from_and_to => (
        required => true,
        isa      => Tuple->of(PositiveOrZeroInt, PositiveOrZeroInt)->where(sub {
            $_->[0] <= $_->[1];
        }),

        # Provide `from` and `to` methods to fetch values
        handles_via => 'Array',
        handles => {
            'from' => [ get => 0 ],
            'to'   => [ get => 1 ],
        },
    );
    
    # Allow `from` and `to` to be separate in the constructor
    method BUILDARGS {
        my %args = ( @_==1 ? %{$_[0]} : @_ );
        $args{from_and_to} ||= [ $args{from}, $args{to} ];
        \%args;
    }
}

Foo->new(from => 0, to => 1); # ok
Foo->new(from => 1, to => 0); # "must be from ≤ to at …&quot;

不过，我不会在类型检查中这样做。如果属性是只读的，或者它是读写的，我会在BUILD（不，不是）中执行此操作。BUILDARGStrigger