I'm learning how to use typeclasses in Haskell.
Consider the following implementation of a typeclass T with a type constrained class function f.
class T t where
f :: (Eq u) => t -> u
data T_Impl = T_Impl_Bool Bool | T_Impl_Int Int | T_Impl_Float Float
instance T T_Impl where
f (T_Impl_Bool x) = x
f (T_Impl_Int x) = x
f (T_Impl_Float x) = x
When I load this into GHCI 7.10.2, I get the following error:
Couldn't match expected type ‘u’ with actual type ‘Float’
‘u’ is a rigid type variable bound by
the type signature for f :: Eq u => T_Impl -> u
at generics.hs:6:5
Relevant bindings include
f :: T_Impl -> u (bound at generics.hs:6:5)
In the expression: x
In an equation for ‘f’: f (T_Impl_Float x) = x
What am I doing/understanding wrong? It seems reasonable to me that one would want to specialize a typeclass in an instance by providing an accompaning data constructor and function implementation. The part
Couldn't match expected type 'u' with actual type 'Float'
is especially confusing. Why does u not match Float if u only has the constraint that it must qualify as an Eq type (Floats do that afaik)?
The signature
f :: (Eq u) => t -> u
means that the caller can pick t and u as wanted, with the only burden of ensuring that u is of class Eq (and t of class T -- in class methods there's an implicit T t constraint).
It does not mean that the implementation can choose any u.
So, the caller can use f in any of these ways: (with t in class T)
f :: t -> Bool
f :: t -> Char
f :: t -> Int
...
The compiler is complaining that your implementation is not general enough to cover all these cases.
Couldn't match expected type ‘u’ with actual type ‘Float’
means "You gave me a Float, but you must provide a value of the general type u (where u will be chosen by the caller)"
Chi has already pointed out why your code doesn't compile. But it's not even that typeclasses are the problem; indeed, your example has only one instance, so it might just as well be a normal function rather than a class.
Fundamentally, the problem is that you're trying to do something like
foobar :: Show x => Either Int Bool -> x
foobar (Left x) = x
foobar (Right x) = x
This won't work. It tries to make foobar return a different type depending on the value you feed it at run-time. But in Haskell, all types must be 100% determined at compile-time. So this cannot work.
There are several things you can do, however.
First of all, you can do this:
foo :: Either Int Bool -> String
foo (Left x) = show x
foo (Right x) = show x
In other words, rather than return something showable, actually show it. That means the result type is always String. It means that which version of show gets called will vary at run-time, but that's fine. Code paths can vary at run-time, it's types which cannot.
Another thing you can do is this:
toInt :: Either Int Bool -> Maybe Int
toInt (Left x) = Just x
toInt (Right x) = Nothing
toBool :: Either Int Bool -> Maybe Bool
toBool (Left x) = Nothing
toBool (Right x) = Just x
Again, that works perfectly fine.
There are other things you can do; without knowing why you want this, it's difficult to suggest others.
As a side note, you want to stop thinking about this like it's object oriented programming. It isn't. It requires a new way of thinking. In particular, don't reach for a typeclass unless you really need one. (I realise this particular example may just be a learning exercise to learn about typeclasses of course...)
It's possible to do this:
class Eq u => T t u | t -> u where
f :: t -> u
You need FlexibleContextx+FunctionalDepencencies and MultiParamTypeClasses+FlexibleInstances on call-site. Or to eliminate class and to use data types instead like Gabriel shows here
Related
There's plenty of Q&A about GADTs being better than DatatypeContexts, because GADTs automagically make constraints available in the right places. For example here, here, here. But sometimes it seems I still need an explicit constraint. What's going on? Example adapted from this answer:
{-# LANGUAGE GADTs #-}
import Data.Maybe -- fromJust
data GADTBag a where
MkGADTBag :: Eq a => { unGADTBag :: [a] } -> GADTBag a
baz (MkGADTBag x) (Just y) = x == y
baz2 x y = unGADTBag x == fromJust y
-- unGADTBag :: GADTBag a -> [a] -- inferred, no Eq a
-- baz :: GADTBag a -> Maybe [a] -> Bool -- inferred, no Eq a
-- baz2 :: Eq a => GADTBag a -> Maybe [a] -> Bool -- inferred, with Eq a
Why can't the type for unGADTBag tell us Eq a?
baz and baz2 are morally equivalent, yet have different types. Presumably because unGADTBag has no Eq a, then the constraint can't propagate into any code using unGADTBag.
But with baz2 there's an Eq a constraint hiding inside the GADTBag a. Presumably baz2's Eq a will want a duplicate of the dictionary already there(?)
Is it that potentially a GADT might have many data constructors, each with different (or no) constraints? That's not the case here, or with typical examples for constrained data structures like Bags, Sets, Ordered Lists.
The equivalent for a GADTBag datatype using DatatypeContexts infers baz's type same as baz2.
Bonus question: why can't I get an ordinary ... deriving (Eq) for GADTBag? I can get one with StandaloneDeriving, but it's blimmin obvious, why can't GHC just do it for me?
deriving instance (Eq a) => Eq (GADTBag a)
Is the problem again that there might be other data constructors?
(Code exercised at GHC 8.6.5, if that's relevant.)
Addit: in light of #chi's and #leftroundabout's answers -- neither of which I find convincing. All of these give *** Exception: Prelude.undefined:
*DTContexts> unGADTBag undefined
*DTContexts> unGADTBag $ MkGADTBag undefined
*DTContexts> unGADTBag $ MkGADTBag (undefined :: String)
*DTContexts> unGADTBag $ MkGADTBag (undefined :: [a])
*DTContexts> baz undefined (Just "hello")
*DTContexts> baz (MkGADTBag undefined) (Just "hello")
*DTContexts> baz (MkGADTBag (undefined :: String)) (Just "hello")
*DTContexts> baz2 undefined (Just "hello")
*DTContexts> baz2 (MkGADTBag undefined) (Just "hello")
*DTContexts> baz2 (MkGADTBag (undefined :: String)) (Just "hello")
Whereas these two give the same type error at compile time * Couldn't match expected type ``[Char]'* No instance for (Eq (Int -> Int)) arising from a use of ``MkGADTBag'/ ``baz2' respectively [Edit: my initial Addit gave the wrong expression and wrong error message]:
*DTContexts> baz (MkGADTBag (undefined :: [Int -> Int])) (Just [(+ 1)])
*DTContexts> baz2 (MkGADTBag (undefined :: [Int -> Int])) (Just [(+ 1)])
So baz, baz2 are morally equivalent not just in that they return the same result for the same well-defined arguments; but also in that they exhibit the same behaviour for the same ill-defined arguments. Or they differ only in where the absence of an Eq instance gets reported?
#leftroundabout Before you've actually deconstructed the x value, there's no way of knowing that the MkGADTBag constructor indeed applies.
Yes there is: field label unGADTBag is defined if and only if there's a pattern match on MkGADTBag. (It would maybe be different if there were other constructors for the type -- especially if those also had a label unGADTBag.) Again, being undefined/lazy evaluation doesn't postpone the type-inference.
To be clear, by "[not] convincing" I mean: I can see the behaviour and the inferred types I'm getting. I don't see that laziness or potential undefinedness gets in the way of type inference. How could I expose a difference between baz, baz2 that would explain why they have different types?
Function calls never bring type class constraints in scope, only (strict) pattern matching does.
The comparison
unGADTBag x == fromJust y
is essentially a function call of the form
foo (unGADTBag x) (fromJust y)
where foo requires Eq a. That would morally be provided by unGADTBag x, but that expression is not yet evaluated! Because of laziness, unGADTBag x will be evaluated only when (and if) foo demands its first argument.
So, in order to call foo in this example we need its argument to be evaluated in advance. While Haskell could work like this, it would be a rather surprising semantics, where arguments are evaluated or not depending on whether they provide a type class constraint which is needed. Imagine more general cases like
foo (if cond then unGADTBag x else unGADTBag z) (fromJust y)
What should be evaluated here? unGADTBag x? unGADTBag y? Both? cond as well? It's hard to tell.
Because of these issues, Haskell was designed so that we need to manually require the evaluation of a GADT value like x using pattern matching.
Why can't the type for unGADTBag tell us Eq a?
Before you've actually deconstructed the x value, there's no way of knowing that the MkGADTBag constructor indeed applies. Sure, if it doesn't then you have other problems (bottom), but those might conceivably not surface. Consider
ignore :: a -> b -> b
ignore _ = id
baz2' :: GADTBag a -> Maybe [a] -> Bool
baz2' x y = ignore (unGADTBag x) (y==y)
Note that I could now invoke the function with, say, undefined :: GADTBag (Int->Int). Shouldn't be a problem since the undefined is ignored, right★? Problem is, despite Int->Int not having an Eq instance, I was able to write y==y, which y :: Maybe [Int->Int] can't in fact support.
So, we can't have that only mentioning unGADTBag is enough to spew the Eq a constraint into its surrounding scope. Instead, we must clearly delimit the scope of that constraint to where we've confirmed that the MkGADTBag constructor does apply, and a pattern match accomplishes that.
★If you're annoyed that my argument relies on undefined, note that the same issue arises also when there are multiple constructors which would bring different constraints into scope.
An alternative to a pattern-match that does work is this:
{-# LANGUAGE RankNTypes #-}
withGADTBag :: GADTBag a -> (Eq a => [a] -> b) -> b
withGADTBag (MkGADTBag x) f = f x
baz3 :: GADTBag a -> Maybe [a] -> Bool
baz3 x y = withGADTBag x (== fromJust y)
Response to edits
All of these give *** Exception: Prelude.undefined:
Yes of course they do, because you actually evaluate x == y in your function. So the function can only possibly yield non-⟂ if the inputs have a NF. But that's by no means the case for all functions.
Whereas these two give the same type error at compile time
Of course they do, because you're trying to wrap a value of non-Eq type in the MkGADTBag constructor, which explicitly requires that constraint (and allows you to explicitly unwrap it again!), whereas the GADTBag type doesn't require that constraint. (Which is kind of the whole point about this sort of encapsulation!)
Before you've actually deconstructed the x value, there's no way of knowing that the `MkGADTBag` constructor indeed applies.Yes there is: field label `unGADTBag` is defined if and only if there's a pattern match on `MkGADTBag`.
Arguably, that's the way field labels should work, but they don't, in Haskell. A field label is nothing but a function from the data type to the field type, and a nontotal function at that if there are multiple constructors.Yeah, Haskell records are one of the worst-designed features of the language. I personally tend to use field labels only for big, single-constructor, plain-old-data types (and even then I prefer using not the field labels directly but lenses derived from them).
Anyway though, I don't see how “field label is defined iff there's a pattern match” could even be implemented in a way that would allow your code to work the way you think it should. The compiler would have to insert the step of confirming that the constructor applies (and extracting its GADT-encapsulated constraint) somewhere. But where? In your example it's reasonably obvious, but in general x could inhabit a vast scope with lots of decision branches and you really don't want it to get evaluated in a branch where the constraint isn't actually needed.
Also keep in mind that when we argue with undefined/⟂ it's not just about actually diverging computations, more typically you're worried about computations that would simply take a long time (just, Haskell doesn't actually have a notion of “taking a long time”).
The way to think about this is OutsideIn(X) ... with local assumptions. It's not about undefinedness or lazy evaluation. A pattern match on a GADT constructor is outside, the RHS of the equation is inside. Constraints from the constructor are made available only locally -- that is only inside.
baz (MkGADTBag x) (Just y) = x == y
Has an explicit data constructor MkGADTBag outside, supplying an Eq a. The x == y raises a wanted Eq a locally/inside, which gets discharged from the pattern match. OTOH
baz2 x y = unGADTBag x == fromJust y
Has no explicit data constructor outside, so no context is supplied. unGADTBag has a Eq a, but that is deeper inside the l.h. argument to ==; type inference doesn't go looking deeper inside. It just doesn't. Then in the effective definition for unGADTBag
unGADTBag (MkGADTBag x) = x
there is an Eq a made available from the outside, but it cannot escape from the RHS into the type environment at a usage site for unGADTBag. It just doesn't. Sad!
The best I can see for an explanation is towards the end of the OutsideIn paper, Section 9.7 Is the emphasis on principal types well-justified? (A rhetorical question but my answer would me: of course we must emphasise principal types; type inference could get better principaled under some circumstances.) That last section considers this example
data R a where
RInt :: Int -> R Int
RBool :: Bool -> R Bool
RChar :: Char -> R Char
flop1 (RInt x) = x
there is a third type that is arguably more desirable [for flop1], and that type is R Int -> Int.
flop1's definition is of the same form as unGADTBag, with a constrained to be Int.
flop2 (RInt x) = x
flop2 (RBool x) = x
Unfortunately, ordinary polymorphic types are too weak to express this restriction [that a must be only Int or Bool] and we can only get Ɐa.R a -> a for flop2, which does not rule the application of flop2 to values of type R Char.
So at that point the paper seems to give up trying to refine better principal types:
In conclusion, giving up on some natural principal types in favor of more specialized types that eliminate more pattern match errors at runtime is appealing but does not quite work unless we consider a more expressive syntax of types. Furthermore it is far from obvious how to specify these typings in a high-level declarative specification.
"is appealing". It just doesn't.
I can see a general solution is difficult/impossible. But for use-cases of constrained Bags/Lists/Sets, the specification is:
All data constructors have the same constraint(s) on the datatype's parameters.
All constructors yield the same type (... -> T a or ... -> T [a] or ... -> T Int, etc).
Datatypes with a single constructor satisfy that trivially.
To satisfy the first bullet, for a Set type using a binary balanced tree, there'd be a non-obvious definition for the Nil constructor:
data OrdSet a where
SNode :: Ord a => OrdSet a -> a -> OrdSet a -> OrdSet a
SNil :: Ord a => OrdSet a -- seemingly redundant Ord constraint
Even so, repeating the constraint on every node and every terminal seems wasteful: it's the same constraint all the way down (which is unlike GADTs for EDSL abstract syntax trees); presumably each node carries a copy of exactly the same dictionary.
The best way to ensure same constraint(s) on every constructor could just be prefixing the constraint to the datatype:
data Ord a => OrdSet a where ...
And perhaps the constraint could go 'OutsideOut' to the environment that's accessing the tree.
Another possible approach is to use a PatternSynonym with an explicit signature giving a Required constraint.
pattern EqGADTBag :: Eq a => [a] -> GADTBag a -- that Eq a is the *Required*
pattern EqGADTBag{ unEqGADTBag } = MkGADTBag unEqGADTBag -- without sig infers Eq a only as *Provided*
That is, without that explicit sig:
*> :i EqGADTBag
pattern EqGADTBag :: () => Eq a => [a] -> GADTBag a
The () => Eq a => ... shows Eq a is Provided, arising from the GADT constructor.
Now we get both inferred baz, baz2 :: Eq a => GADTBag a -> Maybe [a] -> Bool:
baz (EqGADTBag x) (Just y) = x == y
baz2 x y = unEqGADTBag x == fromJust y
As a curiosity: it's possible to give those equations for baz, baz2 as well as those in the O.P. using the names from the GADT decl. GHC warns of overlapping patterns [correctly]; and does infer the constrained sig for baz.
I wonder if there's a design pattern here? Don't put constraints on the data constructor -- that is, don't make it a GADT. Instead declare a 'shadow' PatternSynonym with the Required/Provided constraints.
You can capture the constraint in a fold function, (Eq a => ..) says you can assume Eq a but only within the function next (which is defined after a pattern match). If you instantiate next as = fromJust maybe == as it uses this constraint to witness equality
-- local constraint
-- |
-- vvvvvvvvvvvvvvvvvv
foldGADTBag :: (Eq a => [a] -> res) -> GADTBag a -> res
foldGADTBag next (MkGADTBag as) = next as
baz3 :: GADTBag a -> Maybe [a] -> Bool
baz3 gadtBag maybe = foldGADTBag (fromJust maybe ==) gadtBag
type Ty :: Type -> Type
data Ty a where
TyInt :: Int -> Ty Int
TyUnit :: Ty ()
-- locally assume Int locally assume unit
-- | |
-- vvvvvvvvvvvvvvvvvvv vvvvvvvvvvvvv
foldTy :: (a ~ Int => a -> res) -> (a ~ () => res) -> (Ty a -> res)
foldTy int unit (TyInt i) = int i
foldTy int unit TyUnit = unit
eval :: Ty a -> a
eval = foldTy id ()
Let's say I have an internal data type, T a, that is used in the signature of exported functions:
module A (f, g) where
newtype T a = MkT { unT :: (Int, a) }
deriving (Functor, Show, Read) -- for internal use
f :: a -> IO (T a)
f a = fmap (\i -> T (i, a)) randomIO
g :: T a -> a
g = snd . unT
What is the effect of not exporting the type constructor T? Does it prevent consumers from meddling with values of type T a? In other words, is there a difference between the export list (f, g) and (f, g, T()) here?
Prevented
The first thing a consumer will see is that the type doesn't appear in Haddock documentation. In the documentation for f and g, the type Twill not be hyperlinked like an exported type. This may prevent a casual reader from discovering T's class instances.
More importantly, a consumer cannot doing anything with T at the type level. Anything that requires writing a type will be impossible. For instance, a consumer cannot write new class instances involving T, or include T in a type family. (I don't think there's a way around this...)
At the value level, however, the main limitation is that a consumer cannot write a type annotation including T:
> :t (f . read) :: Read b => String -> IO (A.T b)
<interactive>:1:39: Not in scope: type constructor or class `A.T'
Not prevented
The restriction on type signatures is not as significant a limitation as it appears. The compiler can still infer such a type:
> :t f . read
f . read :: Read b => String -> IO (A.T b)
Any value expression within the inferrable subset of Haskell may therefore be expressed regardless of the availability of the type constructor T. If, like me, you're addicted to ScopedTypeVariables and extensive annotations, you may be a little surprised by the definition of unT' below.
Furthermore, because typeclass instances have global scope, a consumer can use any available class functions without additional limitation. Depending on the classes involved, this may allow significant manipulation of values of the unexposed type. With classes like Functor, a consumer can also freely manipulate type parameters, because there's an available function of type T a -> T b.
In the example of T, deriving Show of course exposes the "internal" Int, and gives a consumer enough information to hackishly implement unT:
-- :: (Show a, Read a) => T a -> (Int, a)
unT' = (read . strip . show') `asTypeOf` (mkPair . g)
where
strip = reverse . drop 1 . reverse . drop 9
-- :: T a -> String
show' = show `asTypeOf` (mkString . g)
mkPair :: t -> (Int, t)
mkPair = undefined
mkString :: t -> String
mkString = undefined
> :t unT'
unT' :: (Show b, Read b) => A.T b -> (Int, b)
> x <- f "x"
> unT' x
(-29353, "x")
Implementing mkT' with the Read instance is left as an exercise.
Deriving something like Generic will completely explode any idea of containment, but you'd probably expect that.
Prevented?
In the corners of Haskell where type signatures are necessary or where asTypeOf-style tricks don't work, I guess not exporting the type constructor could actually prevent a consumer from doing something they could with the export list (f, g, T()).
Recommendation
Export all type constructors that are used in the type of any value you export. Here, go ahead and include T() in your export list. Leaving it out doesn't accomplish anything other than muddying the documentation. If you want to expose an purely abstract immutable type, use a newtype with a hidden constructor and no class instances.
I am having a problem declaring an instance of the following typeclass. I tried to follow the advice in the error message from the ghci compiler but still cannot get the code to compile. Any help would be appreciated.
class TF p where
valid :: p -> Bool
lequiv :: p -> p -> Bool
instance TF Bool
where
valid = id
lequiv f g = f == g
instance TF p => TF (Bool -> p)
where
valid f = valid (f True) && valid (f False)
lequiv f g = (f True) `lequiv` (g True)
&& (f False) `lequiv` (g False)
The error I am getting is:
Illegal instance declaration for ‘TF (Bool -> p)’
(All instance types must be of the form (T a1 ... an)
where a1 ... an are *distinct type variables*,
and each type variable appears at most once in the instance head.
Use FlexibleInstances if you want to disable this.)
In the instance declaration for ‘TF (Bool -> p)’
The problem here is that you have a type constructor (->) applied to things that aren't type variables. There's a lot of ways you can deal with that:
FlexibleInstances. This relaxes the assumption (made in the early days of Haskell, when it wasn't yet clear how difficult implementing type classes would be). This is not very controversial at all. On the other hand, it doesn't play that well with type inference: your instance will only be chosen when we know that we're supplying something of the shape Bool -> p -- and in particular something that's polymorphic in the first argument will not match that shape. So valid id will not typecheck without further annotations.
TypeFamilies. This gives us (among other things) access to a constraint which demands that two particular types be equal. So with this extension, you could write
instance (bool ~ Bool, TF p) => TF (bool -> p) where ...
Now this matches whenever the thing we're supplying has shape bool -> p -- that is, any function at all -- and only after we have selected this instance does it check (in fact, enforce) that the argument type is Bool. This means valid id will typecheck; on the other hand, it also means you cannot declare instances for any other argument types.
Add a typeclass. In fact, the only thing you really care about is that you can list all the inhabitants of Bool in not too much time. So you could instead declare a typeclass, say, Finite, which you will instantiate at such types, and use it as the constraint on the argument type. Thus:
instance (Finite arg, TF p) => TF (arg -> p) where
valid f = all (valid . f) universe
lequiv f g = all (\x -> f x `lequiv` g x) universe
-- could also spell that lambda "liftA2 lequiv f g"
Then you would want to provide a Finite instance for Bool (which, luckily, is already available for you in the universe package). This is nice because it combines the strengths of the previous two approaches: this instance will be chosen as soon as we know the argument is a function, and you can declare instances for many argument types by adding Finite instances for them.
Still new to Haskell, I have hit a wall with the following:
I am trying to define some type classes to generalize a bunch of functions that use gaussian elimination to solve linear systems of equations.
Given a linear system
M x = k
the type a of the elements m(i,j) \elem M can be different from the type b of x and k. To be able to solve the system, a should be an instance of Num and b should have multiplication/addition operators with b, like in the following:
class MixedRing b where
(.+.) :: b -> b -> b
(.*.) :: (Num a) => b -> a -> b
(./.) :: (Num a) => b -> a -> b
Now, even in the most trivial implementation of these operators, I'll get Could not deduce a ~ Int. a is a rigid type variable errors (Let's forget about ./. which requires Fractional)
data Wrap = W { get :: Int }
instance MixedRing Wrap where
(.+.) w1 w2 = W $ (get w1) + (get w2)
(.*.) w s = W $ ((get w) * s)
I have read several tutorials on type classes but I can find no pointer to what actually goes wrong.
Let us have a look at the type of the implementation that you would have to provide for (.*.) to make Wrap an instance of MixedRing. Substituting Wrap for b in the type of the method yields
(.*.) :: Num a => Wrap -> a -> Wrap
As Wrap is isomorphic to Int and to not have to think about wrapping and unwrapping with Wrap and get, let us reduce our goal to finding an implementation of
(.*.) :: Num a => Int -> a -> Int
(You see that this doesn't make the challenge any easier or harder, don't you?)
Now, observe that such an implementation will need to be able to operate on all types a that happen to be in the type class Num. (This is what a type variable in such a type denotes: universal quantification.) Note: this is not the same (actually, it's the opposite) of saying that your implementation can itself choose what a to operate on); yet that is what you seem to suggest in your question: that your implementation should be allowed to pick Int as a choice for a.
Now, as you want to implement this particular (.*.) in terms of the (*) for values of type Int, we need something of the form
n .*. s = n * f s
with
f :: Num a => a -> Int
I cannot think of a function that converts from an arbitary Num-type a to Int in a meaningful way. I'd therefore say that there is no meaningful way to make Int (and, hence, Wrap) an instance of MixedRing; that is, not such that the instance behaves as you would probably expect it to do.
How about something like:
class (Num a) => MixedRing a b where
(.+.) :: b -> b -> b
(.*.) :: b -> a -> b
(./.) :: b -> a -> b
You'll need the MultiParamTypeClasses extension.
By the way, it seems to me that the mathematical structure you're trying to model is really module, not a ring. With the type variables given above, one says that b is an a-module.
Your implementation is not polymorphic enough.
The rule is, if you write a in the class definition, you can't use a concrete type in the instance. Because the instance must conform to the class and the class promised to accept any a that is Num.
To put it differently: Exactly the class variable is it that must be instantiated with a concrete type in an instance definition.
Have you tried:
data Wrap a = W { get :: a }
Note that once Wrap a is an instance, you can still use it with functions that accept only Wrap Int.
I'm struggling to understand the exists keyword in relation to Haskell type system. As far as I know, there is no such keyword in Haskell by default, but:
There are extensions which add them, in declarations like these data Accum a = exists s. MkAccum s (a -> s -> s) (s -> a)
I've seen a paper about them, and (if I recall correctly) it stated that exists keyword is unnecessary for type system since it can be generalized by forall
But I can't even understand what exists means.
When I say, forall a . a -> Int, it means (in my understanding, the incorrect one, I guess) "for every (type) a, there is a function of a type a -> Int":
myF1 :: forall a . a -> Int
myF1 _ = 123
-- okay, that function (`a -> Int`) does exist for any `a`
-- because we have just defined it
When I say exists a . a -> Int, what can it even mean? "There is at least one type a for which there is a function of a type a -> Int"? Why one would write a statement like that? What the purpose? Semantics? Compiler behavior?
myF2 :: exists a . a -> Int
myF2 _ = 123
-- okay, there is at least one type `a` for which there is such function
-- because, in fact, we have just defined it for any type
-- and there is at least one type...
-- so these two lines are equivalent to the two lines above
Please note it's not intended to be a real code which can compile, just an example of what I'm imagining then I hear about these quantifiers.
P.S. I'm not exactly a total newbie in Haskell (maybe like a second grader), but my Math foundations of these things are lacking.
A use of existential types that I've run into is with my code for mediating a game of Clue.
My mediation code sort of acts like a dealer. It doesn't care what the types of the players are - all it cares about is that all the players implement the hooks given in the Player typeclass.
class Player p m where
-- deal them in to a particular game
dealIn :: TotalPlayers -> PlayerPosition -> [Card] -> StateT p m ()
-- let them know what another player does
notify :: Event -> StateT p m ()
-- ask them to make a suggestion
suggest :: StateT p m (Maybe Scenario)
-- ask them to make an accusation
accuse :: StateT p m (Maybe Scenario)
-- ask them to reveal a card to invalidate a suggestion
reveal :: (PlayerPosition, Scenario) -> StateT p m Card
Now, the dealer could keep a list of players of type Player p m => [p], but that would constrict
all the players to be of the same type.
That's overly constrictive. What if I want to have different kinds of players, each implemented
differently, and run them against each other?
So I use ExistentialTypes to create a wrapper for players:
-- wrapper for storing a player within a given monad
data WpPlayer m = forall p. Player p m => WpPlayer p
Now I can easily keep a heterogenous list of players. The dealer can still easily interact with the
players using the interface specified by the Player typeclass.
Consider the type of the constructor WpPlayer.
WpPlayer :: forall p. Player p m => p -> WpPlayer m
Other than the forall at the front, this is pretty standard haskell. For all types
p that satisfy the contract Player p m, the constructor WpPlayer maps a value of type p
to a value of type WpPlayer m.
The interesting bit comes with a deconstructor:
unWpPlayer (WpPlayer p) = p
What's the type of unWpPlayer? Does this work?
unWpPlayer :: forall p. Player p m => WpPlayer m -> p
No, not really. A bunch of different types p could satisfy the Player p m contract
with a particular type m. And we gave the WpPlayer constructor a particular
type p, so it should return that same type. So we can't use forall.
All we can really say is that there exists some type p, which satisfies the Player p m contract
with the type m.
unWpPlayer :: exists p. Player p m => WpPlayer m -> p
When I say, forall a . a -> Int, it
means (in my understanding, the
incorrect one, I guess) "for every
(type) a, there is a function of a
type a -> Int":
Close, but not quite. It means "for every type a, this function can be considered to have type a -> Int". So a can be specialized to any type of the caller's choosing.
In the "exists" case, we have: "there is some (specific, but unknown) type a such that this function has the type a -> Int". So a must be a specific type, but the caller doesn't know what.
Note that this means that this particular type (exists a. a -> Int) isn't all that interesting - there's no useful way to call that function except to pass a "bottom" value such as undefined or let x = x in x. A more useful signature might be exists a. Foo a => Int -> a. It says that the function returns a specific type a, but you don't get to know what type. But you do know that it is an instance of Foo - so you can do something useful with it despite not knowing its "true" type.
It means precisely "there exists a type a for which I can provide values of the following types in my constructor." Note that this is different from saying "the value of a is Int in my constructor"; in the latter case, I know what the type is, and I could use my own function that takes Ints as arguments to do something else to the values in the data type.
Thus, from the pragmatic perspective, existential types allow you to hide the underlying type in a data structure, forcing the programmer to only use the operations you have defined on it. It represents encapsulation.
It is for this reason that the following type isn't very useful:
data Useless = exists s. Useless s
Because there is nothing I can do to the value (not quite true; I could seq it); I know nothing about its type.
UHC implements the exists keyword. Here's an example from its documentation
x2 :: exists a . (a, a -> Int)
x2 = (3 :: Int, id)
xapp :: (exists b . (b,b -> a)) -> a
xapp (v,f) = f v
x2app = xapp x2
And another:
mkx :: Bool -> exists a . (a, a -> Int)
mkx b = if b then x2 else ('a',ord)
y1 = mkx True -- y1 :: (C_3_225_0_0,C_3_225_0_0 -> Int)
y2 = mkx False -- y2 :: (C_3_245_0_0,C_3_245_0_0 -> Int)
mixy = let (v1,f1) = y1
(v2,f2) = y2
in f1 v2
"mixy causes a type error. However, we can use y1 and y2 perfectly well:"
main :: IO ()
main = do putStrLn (show (xapp y1))
putStrLn (show (xapp y2))
ezyang also blogged well about this: http://blog.ezyang.com/2010/10/existential-type-curry/