RE: [Haskell-cafe] type questions again....

-----Original Message----- From: Luke Palmer [mailto:lrpalmer@gmail.com] Sent: 11 January 2008 17:11 To: Nicholls, Mark Cc: haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] type questions again....

2008/1/11 Nicholls, Mark :

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Can someone explain (in simple terms) what is meant by existential and universal types.

Preferably illustrating it in terms of logic rather than lambda calculus.

Well, I don't know about logic. While they are certainly related to existential and universal types in logic, I don't really see a way to explain them in terms of that.

Universal types are easy, you use them every day in Haskell. Take for example id:

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id :: a -> a

Or better illustrated (using ghc extension):

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id :: forall a. a -> a

That means that for any type a I pick, I can get a value of type a -> a from id.

Yep....it's universal because forall types a.

If you wrap an existential type up in a constructor, not much changes:

If you wrap a what?....should this read existential or universal?

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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?

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applyID :: ID -> (Int,String) -> (Int,String) applyID (ID f) (i,s) = (f i, f s)

Note how I can use f, which is a universal type, on both an Int and a String in the same place.

Yep.

You can also put typeclass constraints on universals. Take the universal type forall a. Num a => a -> a. Among functions that have this type are:

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add1 :: forall a. Num a => a -> a add1 x = x + 1

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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?

Wrapping this up in a constructor:

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newtype NUM = NUM (forall a. Num a => a -> a)

We can create values:

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NUM add1 :: NUM NUM id :: NUM

And use them:

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applyNUM :: NUM -> (Int, Double) -> (Int, Double) applyNUM (NUM f) (i,d) = (f i, f d)

Yep.

Existential types are dual,

Dual? (like a dual basis rather than double?)

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

So when you have:

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negNUM' :: NUM' -> NUM' negNUM' (NUM' n) = NUM' (negate n)

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.... 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"... which in term implies that that there exists the function negate... yes? It's existential...because the word "exists" occurred in the above translation.

So we can perform operations to it which work for any Num type, such as negate, but not things that only work for particular Num types, such as div.

Because the existence of the value implies the existence of a type in the typeclass?

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doubleNUM' :: NUM' -> NUM' doubleNUM' (NUM' n) = NUleM' (n + n)

We can add it to itself, but note:

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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"... I think....

How about this:

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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? Don't know complete guess really.

Luke