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I'm learning Haskell and trying to understand the reasoning behind it's syntax design at the same time. Most of the syntax is beautiful.
But since :: normally is like a type annotation, How is it that this works:
Input: minBound::Int
Output: -2147483648
There is no separate operator: :: is a type annotation in that example. Perhaps the best way to understand this is to consider this code:
main = print (f minBound)
f :: Int -> Int
f = id
This also prints -2147483648. The use of minBound is inferred to be an Int because it is the parameter to f. Once the type has been inferred, the value for that type is known.
Now, back to:
main = print (minBound :: Int)
This works in the same way, except that minBound is known to be an Int because of the type annotation, rather than for some more complex reason. The :: isn't some binary operation; it just directs the compiler that the expression minBound has the type Int. Once again, since the type is known, the value can be determined from the type class.
:: still means "has type" in that example.
There are two ways you can use :: to write down type information. Type declarations, and inline type annotations. Presumably you've been used to seeing type declarations, as in:
plusOne :: Integer -> Integer
plusOne = (+1)
Here the plusOne :: Integer -> Integer line is a separate declaration about the identifier plusOne, informing the compiler what its type should be. It is then actually defined on the following line in another declaration.
The other way you can use :: is that you can embed type information in the middle of any expression. Any expression can be followed by :: and then a type, and it means the same thing as the expression on its own except with the additional constraint that it must have the given type. For example:
foo = ('a', 2) :: (Char, Integer)
bar = ('a', 2 :: Integer)
Note that for foo I attached the entire expression, so it is very little different from having used a separate foo :: (Char, Integer) declaration. bar is more interesting, since I gave a type annotation for just the 2 but used that within a larger expression (for the whole pair). 2 :: Integer is still an expression for the value 2; :: is not an operator that takes 2 as input and computes some result. Indeed if the 2 were already used in a context that requires it to be an Integer then the :: Integer annotation changes nothing at all. But because 2 is normally polymorphic in Haskell (it could fit into a context requiring an Integer, or a Double, or a Complex Float) the type annotation pins down that the type of this particular expression is Integer.
The use is that it avoids you having to restructure your code to have a separate declaration for the expression you want to attach a type to. To do that with my simple example would have required something like this:
two :: Integer
two = 2
baz = ('a', two)
Which adds a relatively large amount of extra code just to have something to attach :: Integer to. It also means when you're reading bar, you have to go read a whole separate definition to know what the second element of the pair is, instead of it being clearly stated right there.
So now we can answer your direct question. :: has no special or particular meaning with the Bounded type class or with minBound in particular. However it's useful with minBound (and other type class methods) because the whole point of type classes is to have overloaded names that do different things depending on the type. So selecting the type you want is useful!
minBound :: Int is just an expression using the value of minBound under the constraint that this particular time minBound is used as an Int, and so the value is -2147483648. As opposed to minBound :: Char which is '\NUL', or minBound :: Bool which is False.
None of those options mean anything different from using minBound where there was already some context requiring it to be an Int, or Char, or Bool; it's just a very quick and simple way of adding that context if there isn't one already.
It's worth being clear that both forms of :: are not operators as such. There's nothing terribly wrong with informally using the word operator for it, but be aware that "operator" has a specific meaning in Haskell; it refers to symbolic function names like +, *, &&, etc. Operators are first-class citizens of Haskell: we can bind them to variables1 and pass them around. For example I can do:
(|+|) = (+)
x = 1 |+| 2
But you cannot do this with ::. It is "hard-wired" into the language, just as the = symbol used for introducing definitions is, or the module Main ( main ) where syntax for module headers. As such there are lots of things that are true about Haskell operators that are not true about ::, so you need to be careful not to confuse yourself or others when you use the word "operator" informally to include ::.
1 Actually an operator is just a particular kind of variable name that is applied by writing it between two arguments instead of before them. The same function can be bound to operator and ordinary variables, even at the same time.
Just to add another example, with Monads you can play a little like this:
import Control.Monad
anyMonad :: (Monad m) => Int -> m Int
anyMonad x = (pure x) >>= (\x -> pure (x*x)) >>= (\x -> pure (x+2))
$> anyMonad 4 :: [Int]
=> [18]
$> anyMonad 4 :: Either a Int
=> Right 18
$> anyMonad 4 :: Maybe Int
=> Just 18
it's a generic example telling you that the functionality may change with the type, another example:
Programming in Haskell by Hutton says:
A type that contains one or more type variables is called polymorphic.
Which is a polymorphic type: a type or a set of types?
Is a polymorphic type with a concrete type substituting its type variable a type?
Is a polymorphic type with different concrete types substituting its type variable considered the same or different types?
Is a polymorphic type with a concrete type substituting its type variable a type?
That's the point, yes. However, you need to be careful. Consider:
id :: a -> a
That's polymorphic. You can substitute a := Int and get Int -> Int, and a := Float -> Float and get (Float -> Float) -> Float -> Float. However, you cannot say a := Maybe and get id :: Maybe -> Maybe. That just doesn't make sense. Instead, we have to require that you can only substitute concrete types like Int and Maybe Float for a, not abstract ones like Maybe. This is handled with the kind system. This is not too important for your question, so I'll just summarize. Int and Float and Maybe Float are all concrete types (that is, they have values), so we say that they have type Type (the type of a type is often called its kind). Maybe is a function that takes a concrete type as an argument and returns a new concrete type, so we say Maybe :: Type -> Type. In the type a -> a, we say the type variable a must have type Type, so now the substitutions a := Int, a := String, etc. are allowed, while stuff like a := Maybe isn't.
Is a polymorphic type with different concrete types substituting its type variable considered the same or different types?
No. Back to a -> a: a := Int gives Int -> Int, but a := Float gives Float -> Float. Not the same.
Which is a polymorphic type: a type or a set of types?
Now that's a loaded question. You can skip to the TL;DR at the end, but the question of "what is a polymorphic type" is actually really confusing in Haskell, so here's a wall of text.
There are two ways to see it. Haskell started with one, then moved to the other, and now we have a ton of old literature referring to the old way, so the syntax of the modern system tries to maintain compatibility. It's a bit of a hot mess. Consider
id x = x
What is the type of id? One point of view is that id :: Int -> Int, and also id :: Float -> Float, and also id :: (Int -> Int) -> Int -> Int, ad infinitum, all simultaneously. This infinite family of types can be summed up with one polymorphic type, id :: a -> a. This point of view gives you the Hindley-Milner type system. This is not how modern GHC Haskell works, but this system is what Haskell was based on at its creation.
In Hindley-Milner, there is a hard line between polymorphic types and monomorphic types, and the union of these two groups gives you "types" in general. It's not really fair to say that, in HM, polymorphic types (in HM jargon, "polytypes") are types. You can't take polytypes as arguments, or return them from functions, or place them in a list. Instead, polytypes are only templates for monotypes. If you squint, in HM, a polymorphic type can be seen as a set of those monotypes that fit the schema.
Modern Haskell is built on System F (plus extensions). In System F,
id = \x -> x -- rewriting the example
is not a complete definition. Therefore we can't even think about giving it a type. Every lambda-bound variable needs a type annotation, but x has no annotation. Worse, we can't even decide on one: \(x :: Int) -> x is just as good as \(x :: Float) -> x. In System F, what we do is we write
id = /\(a :: Type) -> \(x :: a) -> x
using /\ to represent Λ (upper-case lambda) much as we use \ to represent λ.
id is a function taking two arguments. The first argument is a Type, named a. The second argument is an a. The result is also an a. The type signature is:
id :: forall (a :: Type). a -> a
forall is a new kind of function arrow, basically. Note that it provides a binder for a. In HM, when we said id :: a -> a, we didn't really define what a was. It was a fresh, global variable. By convention, more than anything else, that variable is not used anywhere else (otherwise the Generalization rule doesn't apply and everything breaks down). If I had written e.g. inject :: a -> Maybe a, afterwards, the textual occurrences of a would be referring to a new global entity, different from the one in id. In System F, the a in forall a. a -> a actually has scope. It's a "local variable" available only for use underneath that forall. The a in inject :: forall a. a -> Maybe a may or may not be the "same" a; it doesn't matter, because we have actual scoping rules that keep everything from falling apart.
Because System F has hygienic scoping rules for type variables, polymorphic types are allowed to do everything other types can do. You can take them as arguments
runCont :: forall (a :: Type). (forall (r :: Type). (a -> r) -> r) -> a
runCons a f = f a (id a) -- omitting type signatures; you can fill them in
You put them in data constructors
newtype Yoneda f a = Yoneda (forall b. (a -> b) -> f b)
You can place them in polymorphic containers:
type Bool = forall a. a -> a -> a
true, false :: Bool
true a t f = t
false a t f = f
thueMorse :: [Bool]
thueMorse = false : true : true : false : _etc
There's an important difference from HM. In HM, if something has polymorphic type, it also has, simultaneously, an infinity of monomorphic types. In System F, a thing can only have one type. id = /\a -> \(x :: a) -> x has type forall a. a -> a, not Int -> Int or Float -> Float. In order to get an Int -> Int out of id, you have to actually give it an argument: id Int :: Int -> Int, and id Float :: Float -> Float.
Haskell is not System F, however. System F is closer to what GHC calls Core, which is an internal language that GHC compiles Haskell to—basically Haskell without any syntax sugar. Haskell is a Hindley-Milner flavored veneer on top of a System F core. In Haskell, nominally a polymorphic type is a type. They do not act like sets of types. However, polymorphic types are still second class. Haskell doesn't let you actually type forall without -XExplicitForalls. It emulates Hindley-Milner's wonky implicit global variable creation by inserting foralls in certain places. The places where it does so are changed by -XScopedTypeVariables. You can't take polymorphic arguments or have polymorphic fields unless you enable -XRankNTypes. You cannot say things like [forall a. a -> a -> a], nor can you say id (forall a. a -> a -> a) :: (forall a. a -> a -> a) -> (forall a. a -> a -> a)—you must define e.g. newtype Bool = Bool { ifThenElse :: forall a. a -> a -> a } to wrap the polymorphism under something monomorphic. You cannot explicitly give type arguments unless you enable -XTypeApplications, and then you can write id #Int :: Int -> Int. You cannot write type lambdas (/\), period; instead, they are inserted implicitly whenever possible. If you define id :: forall a. a -> a, then you cannot even write id in Haskell. It will always be implicitly expanded to an application, id #_.
TL;DR: In Haskell, a polymorphic type is a type. It's not treated as a set of types, or a rule/schema for types, or whatever. However, due to historical reasons, they are treated as second class citizens. By default, it looks like they are treated as mere sets of types, if you squint a bit. Most restrictions on them can be lifted with suitable language extensions, at which point they look more like "just types". The one remaining big restriction (no impredicative instantiations allowed) is rather fundamental and cannot be erased, but that's fine because there's a workaround.
There is some nuance in the word "type" here. Values have concrete types, which cannot be polymorphic. Expressions, on the other hand, have general types, which can be polymorphic. If you're thinking of types for values, then a polymorphic type can be thought of loosely as defining sets of possible concrete types. (At least first-order polymorphic types! Higher-order polymorphism breaks this intuition.) But that's not always a particularly useful way of thinking, and it's not a sufficient definition. It doesn't capture which sets of types can be described in this way (and related notions like parametricity.)
It's a good observation, though, that the same word, "type", is used in these two related, but different, ways.
EDIT: The answer below turns out not to answer the question. The difference is a subtle mistake in terminology: types like Maybe and [] are higher-kinded, whereas types like forall a. a -> a and forall a. Maybe a are polymorphic. The answer below relates to higher-kinded types, but the question was asked about polymorphic types. I’m still leaving this answer up in case it helps anyone else, but I realise now it’s not really an answer to the question.
I would argue that a polymorphic higher-kinded type is closer to a set of types. For instance, you could see Maybe as the set {Maybe Int, Maybe Bool, …}.
However, strictly speaking, this is a bit misleading. To address this in more detail, we need to learn about kinds. Similarly to how types describe values, we say that kinds describe types. The idea is:
A concrete type (that is, one which has values) has a kind of *. Examples include Bool, Char, Int and Maybe String, which all have type *. This is denoted e.g. Bool :: *. Note that functions such as Int -> String also have kind *, as these are concrete types which can contain values such as show!
A type with a type parameter has a kind containing arrows. For instance, in the same way that id :: a -> a, we can say that Maybe :: * -> *, since Maybe takes a concrete type as an argument (such as Int), and produces a concrete type as a result (such as Maybe Int). Something like a -> a also has kind * -> *, since it has one type parameter (a) and produces a concrete result (a -> a). You can get more complex kinds as well: for instance, data Foo f x = FooConstr (f x x) has kind Foo :: (* -> * -> *) -> * -> *. (Can you see why?)
(If the above explanation doesn’t make sense, the Learn You a Haskell book has a great section on kinds as well.)
So now we can answer your questions properly:
Which is a polymorphic higher-kinded type: a type or a set of types?
Neither: a polymorphic higher-kinded type is a type-level function, as indicated by the arrows in its kind. For instance, Maybe :: * -> * is a type-level function which converts e.g. Int → Maybe Int, Bool → Maybe Bool etc.
Is a polymorphic higher-kinded type with a concrete type substituting its type variable a type?
Yes, when your polymorphic higher-kinded type has a kind * -> * (i.e. it has one type parameter, which accepts a concrete type). When you apply a concrete type Conc :: * to a type Poly :: * -> *, it’s just function application, as detailed above, with the result being Poly Conc :: * i.e. a concrete type.
Is a polymorphic higher-kinded type with different concrete types substituting its type variable considered the same or different types?
This question is a bit out of place, as it doesn’t have anything to do with kinds. The answer is definitely no: two types like Maybe Int and Maybe Bool are not the same. Nothing may be a member of both types, but only the former contains a value Just 4, and only the latter contains a value Just False.
On the other hand, it is possible to have two different substitutions where the resulting types are isomorphic. (An isomorphism is where two types are different, but equivalent in some way. For instance, (a, b) and (b, a) are isomorphic, despite being the same type. The formal condition is that two types p,q are isomorphic when you can write two inverse functions p -> q and q -> p.)
One example of this is Const:
data Const a b = Const { getConst :: a }
This type just ignores its second type parameter; as a result, two types like Const Int Char and Const Int Bool are isomorphic. However, they are not the same type: if you make a value of type Const Int Char, but then use it as something of type Const Int Bool, this will result in a type error. This sort of functionality is incredibly useful, as it means you can ‘tag’ a type a using Const a tag, then use the tag as a marker of information on the type level.
Consider these functions
{-# LANGUAGE TypeFamilies #-}
tryMe :: Maybe Int -> Int -> Int
tryMe (Just a) b = a
tryMe Nothing b = b
class Test a where
type TT a
doIt :: TT a -> a -> a
instance Test Int where
type TT Int = Maybe Int
doIt (Just a) b = a
doIt (Nothing) b = b
This works
main = putStrLn $ show $ tryMe (Just 2) 25
This doesn't
main = putStrLn $ show $ doIt (Just 2) 25
{-
• Couldn't match expected type ‘TT a0’ with actual type ‘Maybe a1’
The type variables ‘a0’, ‘a1’ are ambiguous
-}
But then, if I specify the type for the second argument it does work
main = putStrLn $ show $ doIt (Just 2) 25::Int
The type signature for both functions seem to be the same. Why do I need to annotate the second parameter for the type class function? Also, if I annotate only the first parameter to Maybe Int it still doesn't work. Why?
When do I need to cast types in Haskell?
Only in very obscure, pseudo-dependently-typed settings where the compiler can't proove that two types are equal but you know they are; in this case you can unsafeCoerce them. (Which is like C++' reinterpret_cast, i.e. it completely circumvents the type system and just treats a memory location as if it contains the type you've told it. This is very unsafe indeed!)
However, that's not what you're talking about here at all. Adding a local signature like ::Int does not perform any cast, it merely adds a hint to the type checker. That such a hint is needed shouldn't be surprising: you didn't specify anywhere what a is supposed to be; show is polymorphic in its input and doIt polymorphic in its output. But the compiler must know what it is before it can resolve the associated TT; choosing the wrong a might lead to completely different behaviour from the intended.
The more surprising thing is, really, that sometimes you can omit such signatures. The reason this is possible is that Haskell, and more so GHCi, has defaulting rules. When you write e.g. show 3, you again have an ambiguous a type variable, but GHC recognises that the Num constraint can be “naturally” fulfilled by the Integer type, so it just takes that pick.
Defaulting rules are handy when quickly evaluating something at the REPL, but they are fiddly to rely on, hence I recommend you never do it in a proper program.
Now, that doesn't mean you should always add :: Int signatures to any subexpression. It does mean that, as a rule, you should aim for making function arguments always less polymorphic than the results. What I mean by that: any local type variables should, if possible, be deducable from the environment. Then it's sufficient to specify the type of the final end result.
Unfortunately, show violates that condition, because its argument is polymorphic with a variable a that doesn't appear in the result at all. So this is one of the functions where you don't get around having some signature.
All this discussion is fine, but it hasn't yet been stated explicitly that in Haskell numeric literals are polymorphic. You probably knew that, but may not have realized that it has bearing on this question. In the expression
doIt (Just 2) 25
25 does not have type Int, it has type Num a => a — that is, its type is just some numeric type, awaiting extra information to pin it down exactly. And what makes this tricky is that the specific choice might affect the type of the first argument. Thus amalloy's comment
GHC is worried that someone might define an instance Test Integer, in which case the choice of instance will be ambiguous.
When you give that information — which can come from either the argument or the result type (because of the a -> a part of doIt's signature) — by writing either of
doIt (Just 2) (25 :: Int)
doIt (Just 2) 25 :: Int -- N.B. this annotates the type of the whole expression
then the specific instance is known.
Note that you do not need type families to produce this behavior. This is par for the course in typeclass resolution. The following code will produce the same error for the same reason.
class Foo a where
foo :: a -> a
main = print $ foo 42
You might be wondering why this doesn't happen with something like
main = print 42
which is a good question, that leftroundabout has already addressed. It has to do with Haskell's defaulting rules, which are so specialized that I consider them little more than a hack.
With this expression:
putStrLn $ show $ tryMe (Just 2) 25
We've got this starting information to work from:
putStrLn :: String -> IO ()
show :: Show a => a -> String
tryMe :: Maybe Int -> Int -> Int
Just :: b -> Maybe b
2 :: Num c => c
25 :: Num d => d
(where I've used different type variables everywhere, so we can more easily consider them all at once in the same scope)
The job of the type-checker is basically to find types to choose for all of those variables, so and then make sure that the argument and result types line up, and that all the required type class instances exist.
Here we can see that tryMe applied to two arguments is going to be an Int, so a (used as input to show) must be Int. That requires that there is a Show Int instance; indeed there is, so we're done with a.
Similarly tryMe wants a Maybe Int where we have the result of applying Just. So b must be Int, and our use of Just is Int -> Maybe Int.
Just was applied to 2 :: Num c => c. We've decided it must be applied to an Int, so c must be Int. We can do that if we have Num Int, and we do, so c is dealt with.
That leaves 25 :: Num d => d. It's used as the second argument to tryMe, which is expecting an Int, so d must be Int (again discharging the Num constraint).
Then we just have to make sure all the argument and result types line up, which is pretty obvious. This is mostly rehashing the above since we made them line up by choosing the only possible value of the type variables, so I won't get into it in detail.
Now, what's different about this?
putStrLn $ show $ doIt (Just 2) 25
Well, lets look at all the pieces again:
putStrLn :: String -> IO ()
show :: Show a => a -> String
doIt :: Test t => TT t -> t -> t
Just :: b -> Maybe b
2 :: Num c => c
25 :: Num d => d
The input to show is the result of applying doIt to two arguments, so it is t. So we know that a and t are the same type, which means we need Show t, but we don't know what t is yet so we'll have to come back to that.
The result of applying Just is used where we want TT t. So we know that Maybe b must be TT t, and therefore Just :: _b -> TT t. I've written _b using GHC's partial type signature syntax, because this _b is not like the b we had before. When we had Just :: b -> Maybe b we could pick any type we liked for b and Just could have that type. But now we need some specific but unknown type _b such that TT t is Maybe _b. We don't have enough information to know what that type is yet, because without knowing t we don't know which instance's definition of TT t we're using.
The argument of Just is 2 :: Num c => c. So we can tell that c must also be _b, and this also means we're going to need a Num _b instance. But since we don't know what _b is yet we can't check whether there's a Num instance for it. We'll come back to it later.
And finally the 25 :: Num d => d is used where doIt wants a t. Okay, so d is also t, and we need a Num t instance. Again, we still don't know what t is, so we can't check this.
So all up, we've figured out this:
putStrLn :: String -> IO ()
show :: t -> String
doIt :: TT t -> t -> t
Just :: _b -> TT t
2 :: _b
25 :: t
And have also these constraints waiting to be solved:
Test t, Num t, Num _b, Show t, (Maybe _b) ~ (TT t)
(If you haven't seen it before, ~ is how we write a constraint that two type expressions must be the same thing)
And we're stuck. There's nothing further we can figure out here, so GHC is going to report a type error. The particular error message you quoted is complaining that we can't tell that TT t and Maybe _b are the same (it calls the type variables a0 and a1), since we didn't have enough information to select concrete types for them (they are ambiguous).
If we add some extra type signatures for parts of the expression, we can go further. Adding 25 :: Int1 immediately lets us read off that t is Int. Now we can get somewhere! Lets patch that into the constrints we had yet to solve:
Test Int, Num Int, Num _b, Show Int, (Maybe _b) ~ (TT Int)
Num Int and Show Int are obvious and built in. We've got Test Int too, and that gives us the definition TT Int = Maybe Int. So (Maybe _b) ~ (Maybe Int), and therefore _b is Int too, which also allows us to discharge that Num _b constraint (it's Num Int again). And again, it's easy now to verify all the argument and result types match up, since we've filled in all the type variables to concrete types.
But why didn't your other attempt work? Lets go back to as far as we could get with no additional type annotation:
putStrLn :: String -> IO ()
show :: t -> String
doIt :: TT t -> t -> t
Just :: _b -> TT t
2 :: _b
25 :: t
Also needing to solve these constraints:
Test t, Num t, Num _b, Show t, (Maybe _b) ~ (TT t)
Then add Just 2 :: Maybe Int. Since we know that's also Maybe _b and also TT t, this tells us that _b is Int. We also now know we're looking for a Test instance that gives us TT t = Maybe Int. But that doesn't actually determine what t is! It's possible that there could also be:
instance Test Double where
type TT Double = Maybe Int
doIt (Just a) _ = fromIntegral a
doIt Nothing b = b
Now it would be valid to choose t as either Int or Double; either would work fine with your code (since the 25 could also be a Double), but would print different things!
It's tempting to complain that because there's only one instance for t where TT t = Maybe Int that we should choose that one. But the instance selection logic is defined not to guess this way. If you're in a situation where it's possible that another matching instance should exist, but isn't there due to an error in the code (forgot to import the module where it's defined, for example), then it doesn't commit to the only matching instance it can see. It only chooses an instance when it knows no other instance could possibly apply.2
So the "there's only one instance where TT t = Maybe Int" argument doesn't let GHC work backward to settle that t could be Int.
And in general with type families you can only "work forwards"; if you know the type you're applying a type family to you can tell from that what the resulting type should be, but if you know the resulting type this doesn't identify the input type(s). This is often surprising, since ordinary type constructors do let us "work backwards" this way; we used this above to conclude from Maybe _b = Maybe Int that _b = Int. This only works because with new data declarations, applying the type constructor always preserves the argument type in the resulting type (e.g. when we apply Maybe to Int, the resulting type is Maybe Int). The same logic doesn't work with type families, because there could be multiple type family instances mapping to the same type, and even when there isn't there is no requirement that there's an identifiable pattern connecting something in the resulting type to the input type (I could have type TT Char = Maybe (Int -> Double, Bool).
So you'll often find that when you need to add a type annotation, you'll often find that adding one in a place whose type is the result of a type family doesn't work, and you'll need to pin down the input to the type family instead (or something else that is required to be the same type as it).
1 Note that the line you quoted as working in your question main = putStrLn $ show $ doIt (Just 2) 25::Int does not actually work. The :: Int signature binds "as far out as possible", so you're actually claiming that the entire expression putStrLn $ show $ doIt (Just 2) 25 is of type Int, when it must be of type IO (). I'm assuming when you really checked it you put brackets around 25 :: Int, so putStrLn $ show $ doIt (Just 2) (25 :: Int).
2 There are specific rules about what GHC considers "certain knowledge" that there could not possibly be any other matching instances. I won't get into them in detail, but basically when you have instance Constraints a => SomeClass (T a), it has to be able to unambiguously pick an instance only by considering the SomeClass (T a) bit; it can't look at the constraints left of the => arrow.
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.
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