Re: How do subroutines check types?
Ingo Blechschmidt wrote:
How do I have to annotate the type specification in the
declaration of the subroutine to not include the class Foo, but
only allow instances of Foo?
My understanding is that Foo.does(Foo) is false and sub params
are checked with .does(). This automatically excludes class args.
That is you have to explicitly order them:
sub blarb (Foo|Class[Foo] $foo, $arg) # or with CLASS[Foo] ?
{
...; # Do something with instance or class $foo
}
Foo|Class[Foo] is a supertype of Foo and Class[Foo] and as
such allows its subtypes Foo and the Class[Foo] parametric
role instanciation. Note that Foo|Class allows calls like
blarb( Int, "HaHa" ).
Actually I'm not convinced that class Foo should automatically
constitute a (proper) type other than Any. But since .does falls
back to .isa it comes out the same since the only way here to get
a Foo doer is by instanciating the class Foo.
--
TSa (Thomas Sandlaß)
How do subroutines check types?
Hi,
class Foo {...}
Foo.new.isa(Foo); # true
Foo.isa(Foo); # true (see [1])
Foo.does(Class);# true
sub blarb (Foo $foo, $arg) {
...; # Do something with instance $foo
}
blarb Foo.new(...), ...;
# No problem
blarb Foo, ...;
# Problem, as &blarb expects an *instance* of Foo,
# not the class Foo.
How do I have to annotate the type specification in the
declaration of the subroutine to not include the class Foo, but
only allow instances of Foo?
Or is the default way to check the types of arguments
something like the following, in which case my first question
doesn't arise?
if $param ~~ $expected_type and not $param ~~ Class {
# ok
} else {
die "Type error: ...";
}
(But this feels special-casey...)
(And, just curious -- how can I override the default checking
routine?)
--Ingo
[1] http://www.nntp.perl.org/group/perl.perl6.language/0
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