On Jan 11, 2008 5:47 PM, Nicholls, Mark
If you wrap an existential type up in a constructor, not much changes:
If you wrap a what?....should this read existential or universal?
Whoops, right, universal.
newtype ID = ID (forall a. a -> a)
ID can hold any value of type forall a. a -> a; i.e. it can hold any value which exhibits the property that it can give me a value of type a -> a for any type a I choose. In this case the only things ID can hold are id and _|_, because id is the only function that has that type. Here's how I might use it:
It's the only function you've defined the type of....
Id2 :: forall a. a -> a
Now it can hold id2?
Well, that's not what I meant, but yes it can hold id2. What I meant was that, in this case, id2 = _|_ or id2 = id, there are no other possibilities.
id' :: forall a. Num a => a -> a id' = id -- it doesn't have to use the constraint if it doesn't want to
"it doesn't have to use the constraint if it doesn't want to" ?
If id was of type..
Id::forall a. Ord a => a -> a
Then I assume it would complain?
Yes.
but you need to use constructors to use them. I'll write them using GADTs, because I think they're a lot clearer that way:
data NUM' where NUM' :: Num a => a -> NUM'
Look at the type of the constructor NUM'. It has a universal type, meaning whatever type a you pick (as long as it is a Num), you can create a NUM' value with it.
yes
and then it goes wrong...
So the type contained inside the NUM' constructor
?
is called existential (note that NUM' itself is just a regular ADT; NUM' is not existential).
Why existential....see below...I have a guess
Okay, I was being handwavy here. Explaining this will allow me to clear this up. If you take the non-GADT usage of an existential type: data Foo = forall a. Num a => Foo a That is isomorphic to a type: data Foo = Foo (exists a. Num a => a) Except GHC doesn't support a keyword 'exists', and if it did, it would only be able to be used inside constructors like this (in order for inference to be decidable), so why bother? That's what I meant by "the type inside the constructor", Foo is not existential, (exists a. Num a => a) is.
So when you have:
negNUM' :: NUM' -> NUM' negNUM' (NUM' n) = NUM' (negate n)
Here n has an existential type, specifically (exists a. Num a => a).
Here the argument could have been constructed using any numeric type n, so we know very little about it. The only thing we know is that it is of some type a, and that type has a Num instance.
I think one of the harrowing things about Haskell is the practice of overloading data constructors with type names....it confuses the hell out of me....
Yeah that took a little getting used to for me too. But how am I supposed to come up with enough names if I want to name them differently!? That would require too much creativity... :-)
OK so this declaration says that forall x constructed using "NUM' n"...there *exists* a type T s.t. T is a member of type class NUM"...
(you probably meant type class Num here)
which in term implies that that there exists the function negate...
yes?
Huh, I had never thought of it like that, but yes. I just realized that I think of programming in a way quite different than I think of logic. Maybe I should try to have my brain unify them.
doubleNUM' :: NUM' -> NUM' doubleNUM' (NUM' n) = NUleM' (n + n)
We can add it to itself, but note:
addNUM' :: NUM' -> NUM' -> NUM' addNUM' (NUM' a) (NUM' b) = NUM (a + b) -- Illegal!
We can't add them to each other, because the first argument could have been constructed with, say, a Double and the other with a Rational.
But do you see why we're allowed to add it to itself?
We can add it to itself because "+" is of type "a->a->a"...
Yep, so whatever type a n happens to have, it matches both arguments.
How about this:
data Variant where Variant :: a -> Variant
This is a type that can be constructed with any value whatsoever. Looks pretty powerful... but it isn't. Why not?
Eeek.....
Because a could be of any type whatsover?...so how I actually do anything with it?
Right. Luke