I'm trying to store a function type in a definition so I can reuse it, but Haskell doesn't let me do it. A function type is not a data type , nor a class, as far as I understand them. So what am I doing wrong please?
functionType = Int -> Int -> Int -> Int -> Int -> Int -> Int
myfunction :: functionType -- <-- how do I declare this thing in a definition?
myfunction a b c d e f = a*b*c*d*e*f
Type aliases use the type keyword in their declaration; also, as usual for the declaration of new type forms, the newly declared alias must start with an upper case letter*. So:
type FunctionType = Int -> Int -- -> ...
functionValue :: FunctionType
functionValue a = a
* ...or punctuation. Why doesn't the usual "upper-case" punctuation restriction apply? No idea. I never thought about it before trying to write this answer, and now that I have, I find that a bit weird. Perhaps the upper-case restriction on the declaration of new types should be removed!
I have newtype:
newtype Foo = Foo ([Int])
I would like to simply apply Int -> Int function over it like it is possible with fmap.
I thought it will be enough to derive or implement Functor instance, but it requires type of * -> * kind.
Is there some builtin way to make my type partially fmap-able?
https://hackage.haskell.org/package/mono-traversable-1.0.15.1/docs/Data-MonoTraversable.html#t:MonoFunctor
{-# LANGUAGE TypeFamilies #-}
type instance Element Foo = Int
instance MonoFunctor Foo where
-- omap :: (Int -> Int) -> Foo -> Foo
omap = ...
Before getting carried away, you should keep in mind that you're trying to avoid (1) naming and (2) writing the monomorphic function:
mapFoo :: (Int -> Int) -> (Foo -> Foo)
mapFoo f (Foo ints) = Foo (f <$> ints)
If you really don't want to give this function a separate name and want GHC to write the function for you, I think the only sensible way is to re-define your type as a proper functor of kind * -> * with an automatically derived instance:
newtype FooF a = Foo [a] deriving (Functor)
and then define a type alias for the specialization to Int:
type Foo = FooF Int
This is not precisely equivalent to your original definition of Foo. In particular, the following expression in isolation:
Foo [1,2,3]
will have type Num a => FooF a instead of type Foo = FooF Int, so GHC may fail to infer types in all the places it used to. But, this Foo type alias will mostly behave like your original Foo newtype and will allow you to write:
fmap (*5) $ Foo [1,2,3]
and such.
On the other hand, if you want to keep your newtype the same, don't mind writing the function yourself, yet don't want to give that function a separate name, you can't use fmap (at least not without overriding the prelude definition, which kind of defeats the purpose). However, as per #leftroundabout's answer, you can use omap from the mono-traversable package. Because this requires you to define the function yourself in a MonoFunctor Foo instance (e.g., using the same definition as mapFoo above), there's no real point unless you're doing this for a bunch of non-functors besides Foo and want to use a single omap name for all of them or want to write functions that can handle any such MonoFunctor uniformly.
In Scala, I could write the following trait:
trait Consumer[A] {
def apply(a: A): Unit
}
And scala would convert whatever I want to Unit, i.e., it would discard the type. Equivalently, I could have said that apply returns Any and ignore the result.
However, in Haskell, if I defined the type as type Consumer = a -> IO (), I wouldn't be able to pass an Int -> IO Int function, as Int isn't ().
There are two ways I know of solving this issue, but none are satisfactory:
Use Data.Functor.void at the call site to manual change IO a to IO (). This is annoying as an API user.
define type Consumer a b = a -> IO b, but then every time I would want to use Consumer in a signature, I would have to carry the useless type b.
Is there any way to define the Consumer type as a function from a to "IO Any"? As far as I know, Haskell does not support something like exists x. a -> IO x.
Using forall results in the opposite of what I want, e.g.,
type Consumer = forall b. a -> IO b
foo :: Int -> IO Int
foo = undefined
bar :: Consumer Int
bar = foo
results in the error:
• Couldn't match type ‘b’ with ‘Int’
‘b’ is a rigid type variable bound by
the type signature for:
bar :: Consumer Int
Expected type: Int -> IO b
Actual type: Int -> IO Int
• In the expression: foo
In an equation for ‘bar’: bar = foo
• Relevant bindings include
bar :: Int -> IO b
Note that I specifically want Consumer to a be type alias, and not a data constructor, as is described here: Haskell function returning existential type. I wouldn't mind if Consumer were a class if anyone knows how to make that work.
To get an existentially-quantified type in Haskell, you need to write down a data declaration (as opposed to a newtype declaration or a type alias declaration, like you used.).
Here's a Consumer type that fits your purposes:
{-# LANGUAGE ExistentialQuantification #-}
data Consumer input = forall output. Consumer { runDiscardingOutput :: input -> IO output }
And, analogously, here is what your example would look like with the new type:
f :: Int -> IO Int
f = undefined
g :: Consumer Int
g = Consumer f
This doesn't really avoid your concerns about client code needing an extra call, though. (I mean, this is no better than exporting a consumer = Data.Functor.void binding from your library.) Also, it complicates how clients will be able to use a consumer, too:
consumer :: Consumer Int
consumer = Consumer (\x -> return [x])
{- This doesn't typecheck -}
main1 :: IO ()
main1 = runIgnoringOutput consumer 4
{- This doesn't typecheck (!!!) -}
main2 :: IO ()
main2 = void (runIgnoringOutput consumer 4)
{- Only this typechecks :( -}
main3 :: IO ()
main3 =
case consumer of
Consumer f -> Data.Functor.void (f 4)
So it would probably make sense to have a apply function in your library that did the dirty work, just as there was an apply function in the Scala library.
apply :: Consumer a -> a -> IO ()
apply (Consumer f) x = void (f x)
I wouldn't mind if Consumer were a class if anyone knows how to make that work.
You can simulate existential types for classes with an associated type family.
But Haskell doesn't allow ambiguous types in classes without using something like a GADT existential wrapper, so you would still have the type information there somewhere.
{-# LANGUAGE TypeFamilies, MultiParamTypeClasses #-}
class Consumer c a where
type Output c
consume :: c -> a -> IO (Output c)
c is necessary here to allow for the reconstruction of the type of Output c, so it is not strictly an existential. But you can now write
{-# LANGUAGE FlexibleInstances, InstanceSigs #-}
instance Consumer (a -> IO b) a where
type Output (a -> IO b) = b
consume :: (a -> IO b) -> a -> IO b
consume = id
This may not fit your use case, because there will not be a type signature that can express Consumer a in a truly existential way. But it is possible to write
... :: (Consumer c a) => c -> ...
(You could also make use of FunctionalDependencies here to clarify the class somewhat.)
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.
Does haskell-mode or some alternative package offer something akin to the wonderful inferior-haskell-type inside a where clause?
For example, suppose I have
foo = undefined where
bar = complicated
...
it would be super-awesome to be able to place the marker at bar and inspect its type.
This is possible with Scion in combination with Emacs.
C-c C-t shows type of identifier at point. This only works if the current file typechecks, but then it also works for local identifiers. For polymorphic function it will show the type to which they are instantiated, e.g.,
f x = x + (1::Int)
Calling this command on + will print Int -> Int -> Int instead of Num a => a -> a -> a.