I'm new to Purescript. My current learning exercise is to create a tagged union of polymorphic Array and List. I'll use it in a function that finds the length of any Array or List. Here's my attempt:
import Data.List as L
import Data.Array as A
data Collection = CollectionList (forall a. L.List a)
| CollectionArray (forall b. Array b)
colLength :: Collection -> Int
colLength (CollectionList list) = L.length list
colLength (CollectionArray arr) = A.length arr
main :: Effect Unit
main = do
logShow (colLength (CollectionArray [3,5]))
The compiler doesn't like it:
Could not match type Int with type b0
while checking that type Int is at least as general as type b0
while checking that expression 3 has type b0
in value declaration main
where b0 is a rigid type variable
I'm confused by the parts, checking that type Int is at least as general as type b0 and b0 is a rigid type variable. My intention was to allow b to be anything. Not sure what I did to make the compiler put conditions on what b can be.
If you know how, please show the correct way to define a tagged union of polymoric types that'll work in my colLength function.
forall a doesn't mean "any type goes here"
It means that whoever accesses the value, gets to choose what a is, and whoever provides the value has to make sure that the value is of that type. It's a contract between the provider and the consumer.
So when you provide the value CollectionArray [3,5], you have to make it such that it works for all possible a that whoever accesses that value later might choose.
Obviously, there is only one way you can construct such value:
CollectionArray []
What you probably actually meant to do (and I'm guessing here) was to make your collection polymorphic, in the sense that it can contain values of any type, but the type is chosen by whoever creates the collection, and then whoever accesses it later has to deal with that particular type.
To do that, you have to put the type variable on the outside:
data Collection a = CollectionList (L.List a)
| CollectionArray (Array a)
That way, when you create a collection CollectionArray [3,5], it becomes of type Collection Int, and now everywhere you pass it, such as colLength, will have to deal with that Int
This, in turn, can be achieved by making colLength itself generic:
colLength :: forall a. Collection a -> Int
colLength (CollectionList list) = L.length list
colLength (CollectionArray arr) = A.length arr
Now whoever accesses (i.e. calls) colLength itself gets to choose what a is, which works fine, because it's the same place that created the Connection Int in the first place.
Related
Looking into how Either is defined as a functor, I can see that
derive instance functorEither :: Functor (Either a)
which reads to me as "You can map an Either so long as you can map its element.
But either doesn't have just one element. How would this be implemented without derive? Here's what I've tried:
data Either a b = Left a | Right b
instance functorEither :: Functor (Either a)
where
map f (Right b) = Right $ f b
map _ a = a
Of course, the types don't work here:
The Right has this signature: map :: forall a b. (a -> b) -> f a -> f b
The Left however, isn't okay: map :: forall a b. (a -> b) -> f a -> f a
Part of my intuition is saying that Either a b isn't a functor, only Either a is a functor. Which is why map works over Right and ignores Left
That doesn't really give me any intuition for how this is implemented. I still need a way of matching both constructors, don't I?
On the other hand, I think an implementation of map that replaces the inner function with identity is technically law-abiding for functor? The law of composition is met if you just ignore it?
While your proposed definition of the Functor instance indeed fails to compile, it isn't for the reason you say. And it's also "essentially" correct, just not written in a way that will satisfy the compiler.
For convenience, here's your definition again:
data Either a b = Left a | Right b
instance functorEither :: Functor (Either a)
where
map f (Right b) = Right $ f b
map _ a = a
and here's the actual error that you get when trying to compile it:
Could not match type
a02
with type
b1
while trying to match type Either a0 a02
with type Either a0 b1
while checking that expression a
has type Either a0 b1
in value declaration functorEither
where a0 is a rigid type variable
bound at (line 0, column 0 - line 0, column 0)
b1 is a rigid type variable
bound at (line 0, column 0 - line 0, column 0)
a02 is a rigid type variable
bound at (line 0, column 0 - line 0, column 0)
I admit that's a little hard to interpret, if you're not expecting it. But it has to do with the fact that map for Either a needs to have type forall b c. (b -> c) -> Either a b -> Either a c. So the a on the left of map _ a = a has type Either a b, while the one on the right has type Either a c - these are different types (in general), since b and c can be anything, so you can't use the same variable, a, to denote a value of each type.
(This question, although about Haskell rather than Purescript, goes deeper into explanation of exactly this error.)
To fix it, as implied in the question above, you have to explicitly mention that the value you're mapping over is a Left value:
data Either a b = Left a | Right b
instance functorEither :: Functor (Either a)
where
map f (Right b) = Right $ f b
map _ (Left a) = Left a
which is fine because Left a can be interpreted on the left as of type Either a b and on the right as an Either a c.
As for what the instance "does": you are correct that "Either a b isn't a functor, only Either a is a functor" - because a functor must take one type variable, which Either a does but Either a b doesn't. And yes, because the type variable that actually "varies" between Either a b and Either a c is the one that is used in Right, map must only map over the Right values, and leave the Left ones alone - that's the only thing that will satisfy the types needed.
Either a b is often interpreted as representing the result of a computation, where Left values represent failure while Right ones represent success. In this sense it's a slightly "expanded" version of Maybe - the difference is that rather than failure being represented by a single value (Nothing), you get a piece of data (the a type in Either a b) which can tell you information about the error. But the Functor instance works identically to that for Maybe: it maps over any success, and leaves failures alone.
(But there's no logical reason why you can't "map over" the Left values as well. The Bifunctor class is an extension of Functor which can do exactly that.)
I'm having trouble with a Functor instance for a type which is basically just nested Either and Maybe.
data Tuple a b = Tuple a b
data Primitive = String String | Boolean Boolean | Number Number | Null
data JsonValue = Object (Map String JsonValue) | Array (List JsonValue) | Primitive
type Path = List String
data JsonGraphValue = JsonGraphObject (Map String JsonGraphValue) | Atom JsonValue | Ref Path | Error JsonValue | JsonPrimitive Primitive
newtype JsonGraphRecResult a = JsonGraphRecResult (Either String (Tuple (Maybe a) (List Path)))
instance jsonGraphRecResultFunctor :: Functor JsonGraphRecResult where
map f (JsonGraphRecResult (Right (Tuple (Just value) paths))) = JsonGraphRecResult (Right (Tuple (Just (f value)) paths))
map f value = value
I get the following error pointing to the "value" word at the end of the code above.
Could not match type
a1
with type
b0
while trying to match type JsonGraphRecResult a1
with type JsonGraphRecResult b0
while checking that expression value
has type JsonGraphRecResult b0
in value declaration jsonGraphRecResultFunctor
where b0 is a rigid type variable
a1 is a rigid type variable
It's not clear to me why JsonGraphRecResult is any different from the following Blah type which compiles fine:
newtype Blah a = Blah (Maybe a)
instance blahFunctor :: Functor Blah where
map f (Blah (Just x)) = Blah (Just (f x))
map f value = value
The following gist can be pasted directly into the "Try PureScript" online REPL in order to replicate the error.
Figured out the problem. I can't simply return the input value for the map function in the event that Either is Left, because the input value is not the right type. Here's a simplified version of the problem.
-- This is wrong because value is a Functor a, whereas map must return Functor b
map value#Nothing f = value
-- This is right, because even though both sides are Nothing, the right-hand side is a Maybe b vs. Maybe a
map Nothing f = Nothing
You should be able to just derive Functor instance for your type. But before you derive an instance for your final type you should derive it for your Tuple type (or just use Tuple from Data.Tuple ;-)):
data Tuple a b = Tuple a b
derive instance functorTuple :: Functor (Tuple a)
As you can see Functor instance can be defined only for types of kind * -> * so in this case we can map over type which occupies "last position" in Tuple.
In order to derive an instance for your JsonGraphRecResult you have to change order of types in it's internal Tuple to fulfill "last position" requirement:
newtype JsonGraphRecResult a =
JsonGraphRecResult (Either String (Tuple (List Path) (Maybe a)))
derive instance functorJsonGraphRecResult :: Functor JsonGraphRecResult
Here is relevant interactive snippet on trypurescript.org so you can play with this implementation.
There are more type classes for which you can use this deriving mechanism: Ord, Eq, Generic, Newtype...
It is also worth to point out in this context that in Purescript you have additional deriving option which is "newtype deriving". It's syntax is a bit different because it contains newtype keyword after derive - for example:
derive newtype instance someClassMyType :: SomeClass MyType
Newtype deriving is used for newtype "wrappers" for deriving instances of classes which given newtype internal type has already defined instances. In other words when you have newtype T = T a you can derive newtype instances for T for every class which a has instance of.
There is also another strategy - you can also use generic implementations of methods defined for some type classes in Data.Generic.Rep. These implementations can be used for types which are instances of Generic class... but this is whole another story ;-)
You can find more information about deriving in Phil's "24 Days of Purescript" series:
https://github.com/paf31/24-days-of-purescript-2016
So, I am attempting to develop a function that expects a polymorphic function.
I start with and create a class Foo, to represent the constrained type argument in the example function.
module Test
where
class Foo a
Along with a random type for this example...
newtype Bar = Bar Int
In order to pass Bar to anything that has a Foo constraint, it needs to have a typeclass, so I do so:
instance fooBar :: Foo Bar
Everything should be perfectly fine, however, this doesn't work, psc doesn't think the types unify and I don't know why:
example :: forall a. (Foo a) => (a -> Int) -> Int
example a = a (Bar 42)
Error as follows:
Error found:
in module Test
at /Users/arahael/dev/foreign/test.purs line 11, column 16 - line 11, column 22
Could not match type
Bar
with type
a0
while checking that type Bar
is at least as general as type a0
while checking that expression Bar 42
has type a0
in value declaration example
where a0 is a rigid type variable
bound at line 11, column 1 - line 11, column 22
See https://github.com/purescript/documentation/blob/master/errors/TypesDoNotUnify.md for more information,
or to contribute content related to this error.
Compare
(1) example :: (forall a. (Foo a) => a -> Int) -> Int
with
(2) example :: forall a. (Foo a) => (a -> Int) -> Int
The latter essentially places responsibility on you, the caller, to supply as the argument to example a function (your choice!) w/ the signature a -> Int (such that there is a Foo instance for a).
The former, on the other hand, says that you, the caller, don't actually get to pick what this function a -> Int is. Rather, as Norman Ramsey has explained in a Haskell-related stackoverflow answer here, when the forall is "to the left" of the arrow (rather than "above the arrow") the implementation decides what function to supply. The signature in (2) is saying any function a -> Int will do as the argument to example (so long as there is a Foo intance for a).
If (1) just "magically" works, then, it's because the function that unwraps the newtype Bar is being selected by the compiler as perfectly satisfactory, and passed as the argument to example in case (1). After all, the signature of (1) suggests that example should return the same Int values no matter what function a -> Int is given (so long as there is a Foo instance for a). In case (2), the Int values returned by example may very well differ depending upon the specific function a -> Int supplied. The compiler cannot grab any such function that it knows satisfies the type at random. It cannot guess the type of a or a -> Int that you, the caller, actually have in mind.
in insufficiently-polymorphic
the author says about:
def foo[A](fst: List[A], snd: List[A]): List[A]
There are fewer ways we can implement the function. In particular, we
can’t just hard-code some elements in a list, because we have no
ability to manufacture values of an arbitrary type.
I did not understand this, because also in the [Char] version we had no ability to manufacture values of an arbitrary type we had to have them of type [Char] so why are there less ways to implement this?
In the generic version you know that the output list can only contain some arrangement of the elements contained in fst and snd since there is no way to construct new values of some arbitrary type A. In contrast, if you know the output type is Char you can e.g.
def foo(fst: List[Char], snd: List[Char]) = List('a', 'b', 'c')
In addition you cannot use the values contained in the input lists to make decisions which affect the output, since you don't know what they are. You can do this if you know the input type e.g.
def foo(fst: List[Char], snd: List[Char]) = fst match {
case Nil => snd
case 'a'::fs => snd
case _ => fst
}
I'm assuming the author means, that there's no way to construct a non-empty List a but there's a way to construct a List Char, e.g. by using a String literal. You could just ignore the arguments and just return a hard-coded String.
An example of this would be:
foo :: List Char -> List Char -> List Char
foo a b = "Whatever"
You can't construct a value of an arbitrary type a, but you can construct a value of type Char.
This is a simple case of a property called "parametricity" or "free theorem", which applies to every polymorphic function.
An even simpler example is the following:
fun1 :: Int -> Int
fun2 :: forall a. a -> a
fun1 can be anything: successor, predecessor, square, factorial, etc. This is because it can "read" its input, and act accordingly.
fun2 must be the identity function (or loop forever). This because fun2 receives its input, but it can not examine it in any useful way: since it is of an abstract, unknown type a, no operations can be performed on it. The input is effectively an opaque token. The output of foo2 must be of type a, for which we do not know any construction means -- we can not create a value of type a from nothing. The only option is to take the input a and use it to craft the output a. Hence, fun2 is the identity.
The above parametricity result holds when you have no way to perform tests on the input or the type a. If we, e.g., allowed if x.instanceOf[Int] ..., or if x==null ..., or type casts (in OOP) then we could write fun2 in other ways.
I am currently learning Haskell, so here are a beginner's questions:
What is meant by single type in the text below ?
Is single type a special Haskell term ? Does it mean atomic type here ?
Or does it mean that I can never make a list in Haskell in which I can put both 1 and 'c' ?
I was thinking that a type is a set of values.
So I cannot define a type that contains Chars and Ints ?
What about algebraic data types ?
Something like: data IntOrChar = In Int | Ch Char ? (I guess that should work but I am confused what the author meant by that sentence.)
Btw, is that the only way to make a list in Haskell in which I can put both Ints and Chars? Or is there a more tricky way ?
A Scala analogy: in Scala it would be possible to write implicit conversions to a type that represents both Ints and Chars (like IntOrChar) and then it would be possible to put seemlessly Ints and Chars into List[IntOrChar], is that not possible with Haskell ? Do I always have to explicitly wrap every Int or Char into IntOrChar if I want to put them into a list of IntOrChar ?
From Gentle Intro to Haskell:
Haskell also incorporates polymorphic types---types that are
universally quantified in some way over all types. Polymorphic type
expressions essentially describe families of types. For example,
(forall a)[a] is the family of types consisting of, for every type a,
the type of lists of a. Lists of integers (e.g. [1,2,3]), lists of
characters (['a','b','c']), even lists of lists of integers, etc., are
all members of this family. (Note, however, that [2,'b'] is not a
valid example, since there is no single type that contains both 2 and
'b'.)
Short answer.
In Haskell there are no implicit conversions. Also there are no union types - only disjoint unions(which are algebraic data types). So you can only write:
someList :: [IntOrChar]
someList = [In 1, Ch 'c']
Longer and certainly not gentle answer.
Note: This is a technique that's very rarely used. If you need it you're probably overcomplicating your API.
There are however existential types.
{-# LANGUAGE ExistentialQuantification, RankNTypes #-}
class IntOrChar a where
intOrChar :: a -> Either Int Char
instance IntOrChar Int where
intOrChar = Left
instance IntOrChar Char where
intOrChar = Right
data List = Nil
| forall a. (IntOrChar a) => Cons a List
someList :: List
someList = (1 :: Int) `Cons` ('c' `Cons` Nil)
Here I have created a typeclass IntOrChar with only function intOrChar. This way you can convert anything of type forall a. (IntOrChar a) => a to Either Int Char.
And also a special kind of list that uses existential type in its second constructor.
Here type variable a is bound(with forall) at the constructor scope. Therefore every time
you use Cons you can pass anything of type forall a. (IntOrChar a) => a as a first argument. Consequently during a destruction(i.e. pattern matching) the first argument will
still be forall a. (IntOrChar a) => a. The only thing you can do with it is either pass it on or call intOrChar on it and convert it to Either Int Char.
withHead :: (forall a. (IntOrChar a) => a -> b) -> List -> Maybe b
withHead f Nil = Nothing
withHead f (Cons x _) = Just (f x)
intOrCharToString :: (IntOrChar a) => a -> String
intOrCharToString x =
case intOrChar of
Left i -> show i
Right c -> show c
someListHeadString :: Maybe String
someListHeadString = withHead intOrCharToString someList
Again note that you cannot write
{- Wont compile
safeHead :: IntOrChar a => List -> Maybe a
safeHead Nil = Nothing
safeHead (Cons x _) = Just x
-}
-- This will
safeHead2 :: List -> Maybe (Either Int Char)
safeHead2 Nil = Nothing
safeHead2 (Cons x _) = Just (intOrChar x)
safeHead will not work because you want a type of IntOrChar a => Maybe a with a bound at safeHead scope and Just x will have a type of IntOrChar a1 => Maybe a1 with a1 bound at Cons scope.
In Scala there are types that include both Int and Char such as AnyVal and Any, which are both supertypes of Char and Int. In Haskell there is no such hierarchy, and all the basic types are disjoint.
You can create your own union types which describe the concept of 'either an Int or a Char (or you could use the built-in Either type), but there are no implicit conversions in Haskell to transparently convert an Int into an IntOrChar.
You could emulate the concept of 'Any' using existential types:
data AnyBox = forall a. (Show a, Hashable a) => AB a
heteroList :: [AnyBox]
heteroList = [AB (1::Int), AB 'b']
showWithHash :: AnyBox -> String
showWithHash (AB v) = show v ++ " - " ++ (show . hash) v
let strs = map showWithHash heteroList
Be aware that this pattern is discouraged however.
I think that the distinction that is being made here is that your algebraic data type IntOrChar is a "tagged union" - that is, when you have a value of type IntOrChar you will know if it is an Int or a Char.
By comparison consider this anonymous union definition (in C):
typedef union { char c; int i; } intorchar;
If you are given a value of type intorchar you don't know (apriori) which selector is valid. That's why most of the time the union constructor is used in conjunction with a struct to form a tagged-union construction:
typedef struct {
int tag;
union { char c; int i; } intorchar_u
} IntOrChar;
Here the tag field encodes which selector of the union is valid.
The other major use of the union constructor is to overlay two structures to get an efficient mapping between sub-structures. For example, this union is one way to efficiently access the individual bytes of a int (assuming 8-bit chars and 32-bit ints):
union { char b[4]; int i }
Now, to illustrate the main difference between "tagged unions" and "anonymous unions" consider how you go about defining a function on these types.
To define a function on an IntOrChar value (the tagged union) I claim you need to supply two functions - one which takes an Int (in the case that the value is an Int) and one which takes a Char (in case the value is a Char). Since the value is tagged with its type, it knows which of the two functions it should use.
If we let F(a,b) denote the set of functions from type a to type b, we have:
F(IntOrChar,b) = F(Int,b) \times F(Char,b)
where \times denotes the cross product.
As for the anonymous union intorchar, since a value doesn't encode anything bout its type the only functions which can be applied are those which are valid for both Int and Char values, i.e.:
F(intorchar,b) = F(Int,b) \cap F(Char,b)
where \cap denotes intersection.
In Haskell there is only one function (to my knowledge) which can be applied to both integers and chars, namely the identity function. So there's not much you could do with a list like [2, 'b'] in Haskell. In other languages this intersection may not be empty, and then constructions like this make more sense.
To summarize, you can have integers and characters in the same list if you create a tagged-union, and in that case you have to tag each of the values which will make you list look like:
[ I 2, C 'b', ... ]
If you don't tag your values then you are creating something akin to an anonymous union, but since there aren't any (useful) functions which can be applied to both integers and chars there's not really anything you can do with that kind of union.