How do phantom types work with newtype? - haskell

My understanding of newtypes is that they are compiled out by GHC. However, this can't be the whole story because phantom types can hold information.
From here:
you can wrap [a type] in a newtype and it'll be considered distinct to the type-checker, but identical at runtime. You can then use all sorts of deep trickery like phantom or recursive types without worrying about GHC shuffling buckets of bytes for no reason.
For example, imagine a newtype representing arithmetic modulo q:
newtype Zq q = Zq Int
class Modulus q where
getModulus :: q -> Int
addZq :: (Modulus q) => Zq q -> Zq q -> Zq q
addZq (Zq a) (Zq b) = Zq $ (a+b) `mod` (getModulus (undefined :: q))
addZq can't be compiled down to
addZq :: Int -> Int -> Int
so in what sense is the newtype compiled out, and where does the phantom type information get stored?

The thing to keep in mind is that you don't "compile down" to Haskell; you compile down to some other, more explicit language -- in GHC's case, the next well-known step down is core. And although you can't compile addZq down to something of type Int -> Int -> Int in Core, you can compile it down to something whose type you might write as Modulus q => Int -> Int -> Int. In this more explicit language, => has a different meaning than in Haskell; in this language, c => t is the type of a function which takes evidence (in this case, a class dictionary) for the claim c and produces something of type t. So Modulus q => Int -> Int -> Int is roughly the same as (q -> Int) -> Int -> Int -> Int, and addZq certainly can be given that type, even in Haskell.

When newtypes are said to be compiled out by GHC, this means only means runtime behavior. Specifically, they are guaranteed not to be slower than the same code without the newtype. But they still carry on the type information throughout the compilation process, including when used as phantom types.
In other words, a newtype is compile-time information only, and that's good enough for phantom types to work.

Related

differences: GADT, data family, data family that is a GADT

What/why are the differences between those three? Is a GADT (and regular data types) just a shorthand for a data family? Specifically what's the difference between:
data GADT a where
MkGADT :: Int -> GADT Int
data family FGADT a
data instance FGADT a where -- note not FGADT Int
MkFGADT :: Int -> FGADT Int
data family DF a
data instance DF Int where -- using GADT syntax, but not a GADT
MkDF :: Int -> DF Int
(Are those examples over-simplified, so I'm not seeing the subtleties of the differences?)
Data families are extensible, but GADTs are not. OTOH data family instances must not overlap. So I couldn't declare another instance/any other constructors for FGADT; just like I can't declare any other constructors for GADT. I can declare other instances for DF.
With pattern matching on those constructors, the rhs of the equation does 'know' that the payload is Int.
For class instances (I was surprised to find) I can write overlapping instances to consume GADTs:
instance C (GADT a) ...
instance {-# OVERLAPPING #-} C (GADT Int) ...
and similarly for (FGADT a), (FGADT Int). But not for (DF a): it must be for (DF Int) -- that makes sense; there's no data instance DF a, and if there were it would overlap.
ADDIT: to clarify #kabuhr's answer (thank you)
contrary to what I think you're claiming in part of your question, for a plain data family, matching on a constructor does not perform any inference
These types are tricky, so I expect I'd need explicit signatures to work with them. In that case the plain data family is easiest
inferDF (MkDF x) = x -- works without a signature
The inferred type inferDF :: DF Int -> Int makes sense. Giving it a signature inferDF :: DF a -> a doesn't make sense: there is no declaration for a data instance DF a .... Similarly with foodouble :: Foo Char a -> a there is no data instance Foo Char a ....
GADTs are awkward, I already know. So neither of these work without an explicit signature
inferGADT (MkGADT x) = x
inferFGADT (MkFGADT x) = x
Mysterious "untouchable" message, as you say. What I meant in my "matching on those constructors" comment was: the compiler 'knows' on rhs of an equation that the payload is Int (for all three constructors), so you'd better get any signatures consistent with that.
Then I'm thinking data GADT a where ... is as if data instance GADT a where .... I can give a signature inferGADT :: GADT a -> a or inferGADT :: GADT Int -> Int (likewise for inferFGADT). That makes sense: there is a data instance GADT a ... or I can give a signature at a more specific type.
So in some ways data families are generalisations of GADTs. I also see as you say
So, in some ways, GADTs are generalizations of data families.
Hmm. (The reason behind the question is that GHC Haskell has got to the stage of feature bloat: there's too many similar-but-different extensions. I was trying to prune it down to a smaller number of underlying abstractions. Then #HTNW's approach of explaining in terms of yet further extensions is opposite to what would help a learner. IMO existentials in data types should be chucked out: use GADTs instead. PatternSynonyms should be explained in terms of data types and mapping functions between them, not the other way round. Oh, and there's some DataKinds stuff, which I skipped over on first reading.)
As a start, you should think of a data family as a collection of independent ADTs that happen to be indexed by a type, while a GADT is a single data type with an inferrable type parameter where constraints on that parameter (typically, equality constraints like a ~ Int) can be brought into scope by pattern matching.
This means that the biggest difference is that, contrary to what I think you're claiming in part of your question, for a plain data family, matching on a constructor does not perform any inference on the type parameter. In particular, this typechecks:
inferGADT :: GADT a -> a
inferGADT (MkGADT n) = n
but this does not:
inferDF :: DF a -> a
inferDF (MkDF n) = n
and without type signatures, the first would fail to type check (with a mysterious "untouchable" message) while the second would be inferred as DF Int -> Int.
The situation becomes quite a bit more confusing for something like your FGADT type that combines data families with GADTs, and I confess I haven't really thought about how this works in detail. But, as an interesting example, consider:
data family Foo a b
data instance Foo Int a where
Bar :: Double -> Foo Int Double
Baz :: String -> Foo Int String
data instance Foo Char Double where
Quux :: Double -> Foo Char Double
data instance Foo Char String where
Zlorf :: String -> Foo Char String
In this case, Foo Int a is a GADT with an inferrable a parameter:
fooint :: Foo Int a -> a
fooint (Bar x) = x + 1.0
fooint (Baz x) = x ++ "ish"
but Foo Char a is just a collection of separate ADTs, so this won't typecheck:
foodouble :: Foo Char a -> a
foodouble (Quux x) = x
for the same reason inferDF won't typecheck above.
Now, getting back to your plain DF and GADT types, you can largely emulate DFs just using GADTs. For example, if you have a DF:
data family MyDF a
data instance MyDF Int where
IntLit :: Int -> MyDF Int
IntAdd :: MyDF Int -> MyDF Int -> MyDF Int
data instance MyDF Bool where
Positive :: MyDF Int -> MyDF Bool
you can write it as a GADT just by writing separate blocks of constructors:
data MyGADT a where
-- MyGADT Int
IntLit' :: Int -> MyGADT Int
IntAdd' :: MyGADT Int -> MyGADT Int -> MyGADT Int
-- MyGADT Bool
Positive' :: MyGADT Int -> MyGADT Bool
So, in some ways, GADTs are generalizations of data families. However, a major use case for data families is defining associated data types for classes:
class MyClass a where
data family MyRep a
instance MyClass Int where
data instance MyRep Int = ...
instance MyClass String where
data instance MyRep String = ...
where the "open" nature of data families is needed (and where the pattern-based inference methods of GADTs aren't helpful).
I think the difference becomes clear if we use PatternSynonyms-style type signatures for data constructors. Lets start with Haskell 98
data D a = D a a
You get a pattern type:
pattern D :: forall a. a -> a -> D a
it can be read in two directions. D, in "forward" or expression contexts, says, "forall a, you can give me 2 as and I'll give you a D a". "Backwards", as a pattern, it says, "forall a, you can give me a D a and I'll give you 2 as".
Now, the things you write in a GADT definition are not pattern types. What are they? Lies. Lies lies lies. Give them attention only insofar as the alternative is writing them out manually with ExistentialQuantification. Let's use this one
data GD a where
GD :: Int -> GD Int
You get
-- vv ignore
pattern GD :: forall a. () => (a ~ Int) => Int -> GD a
This says: forall a, you can give me a GD a, and I can give you a proof that a ~ Int, plus an Int.
Important observation: The return/match type of a GADT constructor is always the "data type head". I defined data GD a where ...; I got GD :: forall a. ... GD a. This is also true for Haskell 98 constructors, and also data family constructors, though it's a bit more subtle.
If I have a GD a, and I don't know what a is, I can pass into GD anyway, even though I wrote GD :: Int -> GD Int, which seems to say I can only match it with GD Ints. This is why I say GADT constructors lie. The pattern type never lies. It clearly states that, forall a, I can match a GD a with the GD constructor and get evidence for a ~ Int and a value of Int.
Ok, data familys. Lets not mix them with GADTs yet.
data Nat = Z | S Nat
data Vect (n :: Nat) (a :: Type) :: Type where
VNil :: Vect Z a
VCons :: a -> Vect n a -> Vect (S n) a -- try finding the pattern types for these btw
data family Rect (ns :: [Nat]) (a :: Type) :: Type
newtype instance Rect '[] a = RectNil a
newtype instance Rect (n : ns) a = RectCons (Vect n (Rect ns a))
There are actually two data type heads now. As #K.A.Buhr says, the different data instances act like different data types that just happen to share a name. The pattern types are
pattern RectNil :: forall a. a -> Rect '[] a
pattern RectCons :: forall n ns a. Vect n (Rect ns a) -> Rect (n : ns) a
If I have a Rect ns a, and I don't know what ns is, I cannot match on it. RectNil only takes Rect '[] as, RectCons only takes Rect (n : ns) as. You might ask: "why would I want a reduction in power?" #KABuhr has given one: GADTs are closed (and for good reason; stay tuned), families are open. This doesn't hold in Rect's case, as these instances already fill up the entire [Nat] * Type space. The reason is actually newtype.
Here's a GADT RectG:
data RectG :: [Nat] -> Type -> Type where
RectGN :: a -> RectG '[] a
RectGC :: Vect n (RectG ns a) -> RectG (n : ns) a
I get
-- it's fine if you don't get these
pattern RectGN :: forall ns a. () => (ns ~ '[]) => a -> RectG ns a
pattern RectGC :: forall ns' a. forall n ns. (ns' ~ (n : ns)) =>
Vect n (RectG ns a) -> RectG ns' a
-- just note that they both have the same matched type
-- which means there needs to be a runtime distinguishment
If I have a RectG ns a and don't know what ns is, I can still match on it just fine. The compiler has to preserve this information with a data constructor. So, if I had a RectG [1000, 1000] Int, I would incur an overhead of one million RectGN constructors that all "preserve" the same "information". Rect [1000, 1000] Int is fine, though, as I do not have the ability to match and tell whether a Rect is RectNil or RectCons. This allows the constructor to be newtype, as it holds no information. I would instead use a different GADT, somewhat like
data SingListNat :: [Nat] -> Type where
SLNN :: SingListNat '[]
SLNCZ :: SingListNat ns -> SingListNat (Z : ns)
SLNCS :: SingListNat (n : ns) -> SingListNat (S n : ns)
that stores the dimensions of a Rect in O(sum ns) space instead of O(product ns) space (I think those are right). This is also why GADTs are closed and families are open. A GADT is just like a normal data type except it has equality evidence and existentials. It doesn't make sense to add constructors to a GADT any more than it makes sense to add constructors to a Haskell 98 type, because any code that doesn't know about one of the constructors is in for a very bad time. It's fine for families though, because, as you noticed, once you define a branch of a family, you cannot add more constructors in that branch. Once you know what branch you're in, you know the constructors, and no one can break that. You're not allowed to use any constructors if you don't know which branch to use.
Your examples don't really mix GADTs and data families. Pattern types are nifty in that they normalize away superficial differences in data definitions, so let's take a look.
data family FGADT a
data instance FGADT a where
MkFGADT :: Int -> FGADT Int
Gives you
pattern MkFGADT :: forall a. () => (a ~ Int) => Int -> FGADT a
-- no different from a GADT; data family does nothing
But
data family DF a
data instance DF Int where
MkDF :: Int -> DF Int
gives
pattern MkDF :: Int -> DF Int
-- GADT syntax did nothing
Here's a proper mixing
data family Map k :: Type -> Type
data instance Map Word8 :: Type -> Type where
MW8BitSet :: BitSet8 -> Map Word8 Bool
MW8General :: GeneralMap Word8 a -> Map Word8 a
Which gives patterns
pattern MW8BitSet :: forall a. () => (a ~ Bool) => BitSet8 -> Map Word8 a
pattern MW8General :: forall a. GeneralMap Word8 a -> Map Word8 a
If I have a Map k v and I don't know what k is, I can't match it against MW8General or MW8BitSet, because those only want Map Word8s. This is the data family's influence. If I have a Map Word8 v and I don't know what v is, matching on the constructors can reveal to me whether it's known to be Bool or is something else.

Defining a Function for Multiple Types

How is a function defined for different types in Haskell?
Given
func :: Integral a => a -> a
func x = x
func' :: (RealFrac a , Integral b) => a -> b
func' x = truncate x
How could they be combined into one function with the signature
func :: (SomeClassForBoth a, Integral b) => a -> b
With a typeclass.
class TowardsZero a where towardsZero :: Integral b => a -> b
instance TowardsZero Int where towardsZero = fromIntegral
instance TowardsZero Double where towardsZero = truncate
-- and so on
Possibly a class with an associated type family constraint is closer to what you wrote (though perhaps not closer to what you had in mind):
{-# LANGUAGE TypeFamilies #-}
import GHC.Exts
class TowardsZero a where
type RetCon a b :: Constraint
towardsZero :: RetCon a b => a -> b
instance TowardsZero Int where
type RetCon Int b = Int ~ b
towardsZero = id
instance TowardsZero Double where
type RetCon Double b = Integral b
towardsZero = truncate
-- and so on
This is known as ad hoc polymorphism, where you execute different code depending on the type. The way this is done in Haskell is using typeclasses. The most direct way is to define a new class
class Truncable a where
trunc :: Integral b => a -> b
And then you can define several concrete instances.
instance Truncable Integer where trunc = fromInteger
instance Truncable Double where trunc = truncate
This is unsatisfying because it requires an instance for each concrete type, when there are really only two families of identical-looking instances. Unfortunately, this is one of the cases where it is hard to reduce boilerplate, for technical reasons (being able to define "instance families" like this interferes with the open-world assumption of typeclasses, among other difficulties with type inference). As a hint of the complexity, note that your definition assumes that there is no type that is both RealFrac and Integral, but this is not guaranteed -- which implementation should we pick in this case?
There is another issue with this typeclass solution, which is that the Integral version doesn't have the type
trunc :: Integral a => a -> a
as you specified, but rather
trunc :: (Integral a, Integral b) => a -> b
Semantically this is not a problem, as I don't believe it is possible to end up with some polymorphic code where you don't know whether the type you are working with is Integral, but you do need to know that when it is, the result type is the same as the incoming type. That is, I claim that whenever you would need the former rather than the latter signature, you already know enough to replace trunc by id in your source. (It's a gut feeling though, and I would love to be proven wrong, seems like a fun puzzle)
There may be performance implications, however, since you might unnecessarily call fromIntegral to convert a type to itself, and I think the way around this is to use {-# RULES #-} definitions, which is a dark scary bag of complexity that I've never really dug into, so I don't know how hard or easy this is.
I don't recommend this, but you can hack at it with a GADT:
data T a where
T1 :: a -> T a
T2 :: RealFrac a => a -> T b
func :: Integral a => T a -> a
func (T1 x) = x
func (T2 x) = truncate x
The T type says, "Either you already know the type of the value I'm wrapping up, or it's some unknown instance of RealFrac". The T2 constructor existentially quantifies a and packs up a RealFrac dictionary, which we use in the second clause of func to convert from (unknown) a to b. Then, in func, I'm applying an Integral constraint to the a which may or may not be inside the T.

How can an arbitrary Num contain any other numeric type?

I'm just starting with Haskell, and I thought I'd start by making a random image generator. I looked around a bit and found JuicyPixels, which offers a neat function called generateImage. The example that they give doesn't seem to work out of the box.
Their example:
imageCreator :: String -> IO ()
imageCreator path = writePng path $ generateImage pixelRenderer 250 300
where pixelRenderer x y = PixelRGB8 x y 128
when I try this, I get that generateImage expects an Int -> Int -> PixelRGB8 whereas pixelRenderer is of type Pixel8 -> Pixel8 -> PixelRGB8. PixelRGB8 is of type Pixel8 -> Pixel8 -> Pixel8 -> PixelRGB8, so it makes sense that pixelRenderer is doing some type inference to determine that x and y are of type Pixel8. If I define a type signature that asserts that they are of type Int (so the function gets accepted by generateImage, PixelRGB8 complains that it needs Pixel8s, not Ints.
Pixel8 is just a type alias for Word8. After some hair pulling, I discovered that the way to convert an Int to a Word8 is by using fromIntegral.
The type signature for fromIntegral is (Integral a, Num b) => a -> b. It seems to me that the function doesn't actually know what you want to convert it to, so it converts to the very generic Num class. So theoretically, the output of this is a variable of any type that fits the type class Num (correct me if I'm mistaken here--as I understand it, classes are kind of like "interfaces" where types are more like classes/primitives in OOP). If I assign a variable
let n = fromIntegral 5
:t n -- n :: Num b => b
So I'm wondering... what is 'b'? I can use this variable as anything, and it will implicitly cast to any numeric type, as it seems. Not only will it implicitly cast to a Word8, it will implicitly cast to a Pixel8, meaning fromPixel effectively gets turned from (as I understood it) (Integral a, Num b) => a -> b to (Integral a) => a -> Pixel8 depending on context.
Can someone please clarify exactly what's happening here? Why can I use a generic Num as any type that fits Num, both mechanically and "ethically"? I don't understand how the implicit conversion is implemented (if I were to create my own class, I feel like I would need to add explicit conversion functions). I also don't really know why this works; here I can use a pretty unsafe type and convert it implicitly to anything else. (for example, fromIntegral 50000 gets translated to 80 if I implicitly convert it to a Word8)
A common implementation of type classes such as Num is dictionary-passing. Roughly, when the compiler sees something like
f :: Num a => a -> a
f x = x + 2
it transforms it into something like
f :: (Integer -> a, a -> a -> a) -> a -> a
-- ^-- the "dictionary"
f (dictFromInteger, dictPlus) x = dictPlus x (dictFromInteger 2)
The latter basically says: "pass me an implementation for these methods of class Num for your type a, and I will use them to produce a function a -> a for you".
Values such as your n :: Num b => b are no different. They are compiled into things such as
n :: (Integer -> b) -> b
n dictFromInteger = dictFromInteger 5 -- roughly
As you can see, this turns innocent-looking integer literals into functions, which can (and does) impact performance. However, in many circumstances the compiler can realize that the full polymorphic version is not actually needed, and remove all the dictionaries.
For instance, if you write f 3 but f expects Int, the "polymorphic" 3 can be converted at compile time. So type inference can aid the optimization phase (and user-written type annotation can greatly help here). Further, some other optimizations can be triggered manually, e.g. using the GHC SPECIALIZE pragma. Finally, the dreaded monomorphism restriction tries hard to force non-functions to remain non-functions after translation, at the cost of some loss of polymorphism. However, the MR is now being regarded as harmful, since it can cause puzzling type errors in some contexts.

Haskell get type of algebraic parameter

I have a type
class IntegerAsType a where
value :: a -> Integer
data T5
instance IntegerAsType T5 where value _ = 5
newtype (IntegerAsType q) => Zq q = Zq Integer deriving (Eq)
newtype (Num a, IntegerAsType n) => PolyRing a n = PolyRing [a]
I'm trying to make a nice "show" for the PolyRing type. In particular, I want the "show" to print out the type 'a'. Is there a function that returns the type of an algebraic parameter (a 'show' for types)?
The other way I'm trying to do it is using pattern matching, but I'm running into problems with built-in types and the algebraic type.
I want a different result for each of Integer, Int and Zq q.
(toy example:)
test :: (Num a, IntegerAsType q) => a -> a
(Int x) = x+1
(Integer x) = x+2
(Zq x) = x+3
There are at least two different problems here.
1) Int and Integer are not data constructors for the 'Int' and 'Integer' types. Are there data constructors for these types/how do I pattern match with them?
2) Although not shown in my code, Zq IS an instance of Num. The problem I'm getting is:
Ambiguous constraint `IntegerAsType q'
At least one of the forall'd type variables mentioned by the constraint
must be reachable from the type after the '=>'
In the type signature for `test':
test :: (Num a, IntegerAsType q) => a -> a
I kind of see why it is complaining, but I don't know how to get around that.
Thanks
EDIT:
A better example of what I'm trying to do with the test function:
test :: (Num a) => a -> a
test (Integer x) = x+2
test (Int x) = x+1
test (Zq x) = x
Even if we ignore the fact that I can't construct Integers and Ints this way (still want to know how!) this 'test' doesn't compile because:
Could not deduce (a ~ Zq t0) from the context (Num a)
My next try at this function was with the type signature:
test :: (Num a, IntegerAsType q) => a -> a
which leads to the new error
Ambiguous constraint `IntegerAsType q'
At least one of the forall'd type variables mentioned by the constraint
must be reachable from the type after the '=>'
I hope that makes my question a little clearer....
I'm not sure what you're driving at with that test function, but you can do something like this if you like:
{-# LANGUAGE ScopedTypeVariables #-}
class NamedType a where
name :: a -> String
instance NamedType Int where
name _ = "Int"
instance NamedType Integer where
name _ = "Integer"
instance NamedType q => NamedType (Zq q) where
name _ = "Zq (" ++ name (undefined :: q) ++ ")"
I would not be doing my Stack Overflow duty if I did not follow up this answer with a warning: what you are asking for is very, very strange. You are probably doing something in a very unidiomatic way, and will be fighting the language the whole way. I strongly recommend that your next question be a much broader design question, so that we can help guide you to a more idiomatic solution.
Edit
There is another half to your question, namely, how to write a test function that "pattern matches" on the input to check whether it's an Int, an Integer, a Zq type, etc. You provide this suggestive code snippet:
test :: (Num a) => a -> a
test (Integer x) = x+2
test (Int x) = x+1
test (Zq x) = x
There are a couple of things to clear up here.
Haskell has three levels of objects: the value level, the type level, and the kind level. Some examples of things at the value level include "Hello, world!", 42, the function \a -> a, or fix (\xs -> 0:1:zipWith (+) xs (tail xs)). Some examples of things at the type level include Bool, Int, Maybe, Maybe Int, and Monad m => m (). Some examples of things at the kind level include * and (* -> *) -> *.
The levels are in order; value level objects are classified by type level objects, and type level objects are classified by kind level objects. We write the classification relationship using ::, so for example, 32 :: Int or "Hello, world!" :: [Char]. (The kind level isn't too interesting for this discussion, but * classifies types, and arrow kinds classify type constructors. For example, Int :: * and [Int] :: *, but [] :: * -> *.)
Now, one of the most basic properties of Haskell is that each level is completely isolated. You will never see a string like "Hello, world!" in a type; similarly, value-level objects don't pass around or operate on types. Moreover, there are separate namespaces for values and types. Take the example of Maybe:
data Maybe a = Nothing | Just a
This declaration creates a new name Maybe :: * -> * at the type level, and two new names Nothing :: Maybe a and Just :: a -> Maybe a at the value level. One common pattern is to use the same name for a type constructor and for its value constructor, if there's only one; for example, you might see
newtype Wrapped a = Wrapped a
which declares a new name Wrapped :: * -> * at the type level, and simultaneously declares a distinct name Wrapped :: a -> Wrapped a at the value level. Some particularly common (and confusing examples) include (), which is both a value-level object (of type ()) and a type-level object (of kind *), and [], which is both a value-level object (of type [a]) and a type-level object (of kind * -> *). Note that the fact that the value-level and type-level objects happen to be spelled the same in your source is just a coincidence! If you wanted to confuse your readers, you could perfectly well write
newtype Huey a = Louie a
newtype Louie a = Dewey a
newtype Dewey a = Huey a
where none of these three declarations are related to each other at all!
Now, we can finally tackle what goes wrong with test above: Integer and Int are not value constructors, so they can't be used in patterns. Remember -- the value level and type level are isolated, so you can't put type names in value definitions! By now, you might wish you had written test' instead:
test' :: Num a => a -> a
test' (x :: Integer) = x + 2
test' (x :: Int) = x + 1
test' (Zq x :: Zq a) = x
...but alas, it doesn't quite work like that. Value-level things aren't allowed to depend on type-level things. What you can do is to write separate functions at each of the Int, Integer, and Zq a types:
testInteger :: Integer -> Integer
testInteger x = x + 2
testInt :: Int -> Int
testInt x = x + 1
testZq :: Num a => Zq a -> Zq a
testZq (Zq x) = Zq x
Then we can call the appropriate one of these functions when we want to do a test. Since we're in a statically-typed language, exactly one of these functions is going to be applicable to any particular variable.
Now, it's a bit onerous to remember to call the right function, so Haskell offers a slight convenience: you can let the compiler choose one of these functions for you at compile time. This mechanism is the big idea behind classes. It looks like this:
class Testable a where test :: a -> a
instance Testable Integer where test = testInteger
instance Testable Int where test = testInt
instance Num a => Testable (Zq a) where test = testZq
Now, it looks like there's a single function called test which can handle any of Int, Integer, or numeric Zq's -- but in fact there are three functions, and the compiler is transparently choosing one for you. And that's an important insight. The type of test:
test :: Testable a => a -> a
...looks at first blush like it is a function that takes a value that could be any Testable type. But in fact, it's a function that can be specialized to any Testable type -- and then only takes values of that type! This difference explains yet another reason the original test function didn't work. You can't have multiple patterns with variables at different types, because the function only ever works on a single type at a time.
The ideas behind the classes NamedType and Testable above can be generalized a bit; if you do, you get the Typeable class suggested by hammar above.
I think now I've rambled more than enough, and likely confused more things than I've clarified, but leave me a comment saying which parts were unclear, and I'll do my best.
Is there a function that returns the type of an algebraic parameter (a 'show' for types)?
I think Data.Typeable may be what you're looking for.
Prelude> :m + Data.Typeable
Prelude Data.Typeable> typeOf (1 :: Int)
Int
Prelude Data.Typeable> typeOf (1 :: Integer)
Integer
Note that this will not work on any type, just those which have a Typeable instance.
Using the extension DeriveDataTypeable, you can have the compiler automatically derive these for your own types:
{-# LANGUAGE DeriveDataTypeable #-}
import Data.Typeable
data Foo = Bar
deriving Typeable
*Main> typeOf Bar
Main.Foo
I didn't quite get what you're trying to do in the second half of your question, but hopefully this should be of some help.

Writing a function with a universally-quantified return type

If I write
foo :: (Num a) => a
foo = 42
GHC happily accepts it, but if I write
bar :: (Num a) => a
bar = (42 :: Int)
it tells me that the expected type a doesn't match the inferred type Int. I don't quite understand why, since Int is an instance of the class Num that a stands for.
I'm running into this same situation while trying to write a function that, boiled down to the core of the problem, looks roughly like this:
-- Note, Frob is an instance of class Frobbable
getFrobbable :: (Frobbable a) => Frob -> a
getFrobbable x = x
Is it possible to write a function like this? How can I make the result compatible with the type signature?
You are treating typeclass constraints as if they were subtype constraints. This is a common thing to do, but in fact they only coincide in the contravariant case, that is, when concerning function arguments rather than results. The signature:
bar :: (Num a) => a
Means that the caller gets to choose the type a, provided that it is an instance of Num. So here, the caller may choose to call bar :: Int or bar :: Double, they should all work. So bar :: (Num a) => a must be able to construct any kind of number, bar does not know what specific type was chosen.
The contravariant case is exactly the same, it just corresponds with OO programmers' intuition in this case. Eg.
baz :: (Num a) => a -> Bool
Means that the caller gets to choose the type a, again, and again baz does not know what specific type was chosen.
If you need to have the callee choose the result type, just change the signature of the function to reflect this knowledge. Eg.:
bar :: Int
Or in your getFrobbable case, getFrobbable :: Frob -> Frob (which makes the function trivial). Anywhere that a (Frobbable a) constraint occurs, a Frob will satisfy it, so just say Frob instead.
This may seem awkward, but it's really just that information hiding happens at different boundaries in functional programming. There is a way to hide the specific choice, but it is uncommon, and I consider it a mistake in most cases where it is used.
One way to get the effect your sample seems to be going for
I first want to describe a way to accomplish what it appears you're going for. Let's look at your last code sample again:
-- Note, Frob is an instance of class Frobbable
getFrobbable :: (Frobbable a) => Frob -> a
getFrobbable x = x
This is essentially a casting operation. It takes a Frob and simply forgets what it is, retaining only the knowledge that you've got an instance of Frobbable.
There is an idiom in Haskell for accomplishing this. It requires the ExistentialQuantification extension in GHC. Here is some sample code:
{-# LANGUAGE ExistentialQuantification #-}
module Foo where
class Frobbable a where
getInt :: a -> Int
data Frob = Frob Int
instance Frobbable Frob where
getInt (Frob i) = i
data FrobbableWrapper = forall a . Frobbable a => FW a
instance Frobbable FrobbableWrapper where
getInt (FW i) = getInt i
The key part is the FrobbableWrapper data structure. With it, you can write the following version of your getFrobbable casting function:
getFrobbable :: Frobbable a => a -> FrobbableWrapper
getFrobbable x = FW x
This idiom is useful if you want to have a heterogeneous list whose elements share a common typeclass, even though they may not share a common type. For instance, while Frobbable a => [a] wouldn't allow you to mix different instances of Frobbable, the list [FrobbableWrapper] certainly would.
Why the code you posted isn't allowed
Now, why can't you write your casting operation as-is? It's all about what could be accomplished if your original getFrobbable function was permitted to type check.
The equation getFrobbable x = x really should be thought of as an equation. x isn't being modified in any way; thus, neither is its type. This is done for the following reason:
Let's compare getFrobbable to another object. Consider
anonymousFrobbable :: Frobbable a => a
anonymousFrobbable = undefined
(Code involving undefined are a great source of awkward behavior when you want to really push on your intuition.)
Now suppose someone comes along and introduces a data definition and a function like
data Frob2 = Frob2 Int Int
instance Frobbable Frob2 where
getInt (Frob2 x y) = y
useFrobbable :: Frob2 -> [Int]
useFrobbable fb2 = []
If we jump into ghci we can do the following:
*Foo> useFrobbable anonymousFrobbable
[]
No problems: the signature of anonymousFrobbable means "You pick an instance of Frobbable, and I'll pretend I'm of that type."
Now if we tried to use your version of getFrobbable, a call like
useFrobbable (getFrobbable someFrob)
would lead to the following:
someFrob must be of type Frob, since it is being given to getFrobbable.
(getFrobbable someFrob) must be of type Frob2 since it is being given to useFrobbable
But by the equation getFrobbable someFrob = someFrob, we know that getFrobbable someFrob and someFrob have the same type.
Thus the system concludes that Frob and Frob2 are the same type, even though they are not. Hence this reasoning is unsound, which is ultimately the type of rational behind why the version of getFrobbable you posted doesn't type-check.
Also worth noting is that the literal 42 actually stands for fromInteger (42 :: Integer), which really has type (Num a) => a. See the Haskell Report on numeric literals.
A similar mechanism works for floating point numbers and you can get GHC to use overloaded string literals, but for other types there is (to my knowledge) no way to do that.
I am a teensy bit confused as to what you're trying to do, but if you just want the caller's use of the return value to drive instance selection, then that's normal typeclasses at work.
data Frob = Frob Int Float
class FrobGettable a where
getFrobbable :: Frob -> a
instance FrobGettable Int where
getFrobbable (Frob x _) = x
instance FrobGettable Float where
getFrobbable (Frob _ y) = y
instance FrobGettable Char where
getFrobbable _ = '!'
frob = Frob 10 3.14
test1 :: Float
test1 = getFrobbable frob * 1.1
test2 :: Int
test2 = getFrobbable frob `div` 4
test3 = "Hi" ++ [getFrobbable frob]
We can use GHCi to see what we have,
*Main> :t getFrobbable
getFrobbable :: (FrobGettable a) => Frob -> a

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