Is every class not a type in Haskell :
Prelude> :t max
max :: Ord a => a -> a -> a
Prelude> :t Ord
<interactive>:1:1: Not in scope: data constructor ‘Ord’
Prelude>
Why does this not print Ord type signature ?
Okay, there's a couple of things going on here.
First when you write :t Ord you're looking for something called Ord in the value namespace; specifically it would have to be a constructor, since the name starts with a capital letter.
Haskell keeps types and values completely separate; there is no relationship between the name of a type and the names of a type's constructors. Often when there's only one constructor, people will use the same name as the type. An example being data Foo = Foo Int. This declares two new named entities: the type Foo and the constructor Foo :: Int -> Foo.
It's not really a good idea to think of it as just making a type Foo that can be used both in type expressions and to construct Foos. Because also common are declarations like data Maybe a = Nothing | Just a. Here there are 2 different constructors for Maybe a, and Maybe isn't a name of anything at all at the value level.
So just because you've seen Ord in a type expression doesn't mean that there is a name Ord at the value level for you to ask the type of with :t. Even if there were, it wouldn't necessarily be related top the type-level name Ord.
The second point that needs clarifying is that no, classes are not in fact types. A class is a set of types (which all support the interface defined in the class), but it is not a type itself.
In vanilla Haskell type classes are just "extra" things. You can declare them with a class declaration, instantiate them with an instance declaration, and use them in special syntax attached to types (the stuff left of the => arrow) as constraints on type variables. But they don't really interact with the rest of the language, and you cannot use them in the main part of a type signature (the stuff right of the `=> arrow).
However, with the ConstraintKinds extension on, type classes do become ordinary things that exist in the type namespace, just like Maybe. They are still not types in the sense that there can never be any values that have them as types, so you can't use Ord or Ord Int as an argument or return type in a function, or have a [Ord a] or anything like that.
In that they are a bit like type constructors like Maybe. Maybe is a name bound in the type namespace, but it is not a type as such; there are no values whose type is just Maybe, but Maybe can be used as part of an expression defining a type, as in Maybe Int.
If you're not familiar with kinds, probably ignore everything I've said from ConstraintKinds onwards; you'll probably learn about kinds eventually, but they're not a feature you need to know much about as a beginner. If you are, however, what ConstraintKinds does is make a special kind Constraint and have type class constraints (left of the => arrow) just be ordinary type-level things of kind Constraint instead of special purpose syntax. This means that Ord is a type-level thing, and we can ask it's kind with the :k command in GHCI:
Prelude> :k Ord
* -> Constraint
Which makes sense; max had type Ord a => a -> a -> a, so Ord a must have kind Constraint. If Ord can be applied to an ordinary type to yield a constraint, it must have kind * -> Constraint.
Ord isn't a type; it's a typeclass. Typeclasses allow you to associate supported operations with a given type (somewhat similar to interfaces in Java or protocols in Objective-C). A type (e.g. Int) being an "instance" of a typeclass (e.g. Ord) means that the type supports the functions of the Ord typeclass (e.g. compare, <, > etc.).
You can get most info about a typeclass using :i in ghci, which shows you the functions associated with the typeclass and which types are instances of it:
ghci > :i Ord
class Eq a => Ord a where
compare :: a -> a -> Ordering
(<) :: a -> a -> Bool
(>=) :: a -> a -> Bool
(>) :: a -> a -> Bool
(<=) :: a -> a -> Bool
max :: a -> a -> a
min :: a -> a -> a
-- Defined in ‘GHC.Classes’
instance Ord a => Ord (Maybe a) -- Defined in ‘Data.Maybe’
instance (Ord a, Ord b) => Ord (Either a b)
-- Defined in ‘Data.Either’
instance Ord Integer -- Defined in ‘integer-gmp:GHC.Integer.Type’
instance Ord a => Ord [a] -- Defined in ‘GHC.Classes’
...
Ord is not a type, but a typeclass. It does not have a type, but a kind:
Prelude> :k Ord
Ord :: * -> Constraint
Typeclasses are one of the wonderful things about Haskell. Check 'em out :-)
Not quite. You can impose type constraints, so Ord a => a is a type, but Ord a isn't. Ord a => a means "any type a with the constraint that it is an instance of Ord".
The error is because :t expects an expression. When GHCi tries to interpret Ord as an expression, the closest it can get to is a data constructor, since these are the only functions in Haskell that can start with capital letters.
Related
This question already has answers here:
Type Constraints in Data Declaration Haskell
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Closed 3 years ago.
I am trying to make a normal form game solver for game theory, and I'm trying to make it as generic as possible for good practice and for my own convenience. I would like to use the same functions to solve both zero-sum and non-zero-sum games, so I am using the following data type:
data Payoffs = (Num a, Eq a, Ord a) => ZS a
| (Num a, Eq a, Ord a) => NZS (a,a)
However, this is not correct syntax. Is there any way to constrain a so that it must satisfy those type constraints?
Short answer (and probably not the one you need):
To make your code work as is, you need a forall quantifier (for which you need to enable ExistentialQuantification):
{-# LANGUAGE ExistentialQuantification #-}
data Payoffs =
forall a. (Num a, Eq a, Ord a) => ZS a
| forall a. (Num a, Eq a, Ord a) => NZS (a,a)
If you have a type variable in the data constructor (i.e. ZS a), then you have two choices: either that variable has to appear in the type constructor (i.e. data Payoffs a =), or you need to say "I don't care what type it is, as long as it supports these classes" - which is achieved via the forall quantifier.
But this looks kinda useless to me, which suggests that you may be misunderstanding what it means. If you write the above code, every value of your Payoffs type will be able to wrap a value of any type, as long as that type supports Num, Eq, and Ord. One subtle consequence of this is that, if you have two values of Payoffs lying around, they will not necessarily wrap the same type. For example:
let x = ZS (42 :: Int) -- wraps an Int
let y = NZS (2.71 :: Double, 3.14) -- wraps two Doubles
This means that, upon unpacking them, you won't be able to, for example, add them together, because, even though they both implement Num, the compiler doesn't have any proof that they're actually the same type.
What I suspect you actually need is a parametrized type, like this:
data Payoffs a = ZS a | NZS (a, a)
But then, of course, you lose the constraints: anybody can go and create ZS String or something. You can use the GADT syntax (with the GADTs extension) to bring them back:
{-# LANGUAGE GADTs #-}
data Payoffs a where
ZS :: (Num a, Ord a, Eq a) => a -> Payoffs a
NZS :: (Num a, Ord a, Eq a) => (a, a) -> Payoffs a
This notation is equivalent to ZS a | NZS (a, a), except you get to define each constructor with the same syntax as any function - including constraints. A type defined like this won't allow for creating values of type Payoffs a unless a satisfies the constraints.
At the same time, if you have a value of a type like this lying around, you know what type it wraps inside. And this allows you to tell if two Payoffs values wrap the same type or different. And then, if you know that they're the same, you can do things with them using the supported classes, for example:
addPayoffs :: Payoffs a -> Payoffs a -> Payoffs a
addPayoffs (ZS a) (ZS b) = ZS (a + b)
addPayoffs (ZS a) (NZS (x,y)) = NZS (a+x, a+y)
... etc.
When I am in a REPL like GHCI with Prelude, and I write
*> compare 5 7
LT
Why can I call that function (compare) like that directly in the REPL?
I know that compare is defined in typeclass Ord. The typeclass definition for Ord of course shows that it is a subclass of Eq.
Here is my line of reasoning:
5 has type Num a => a, and Num typeclass is not a subclass of Eq.
Also,
Prelude> :t (compare 5)
(compare 5) :: (Num a, Ord a) => a -> Ordering
So, there is an additional constraint imposed here when I apply a numeric type argument. when I call compare 5 7, the types of the arguments are narrowed to something that does have an instance of Ord. I think the narrowing happens to the default concrete type associated with the typeclass: in the case of Num, this is Integer, which has an instance of Real, which has an instance of Ord.
However, coming from a non-functional programming background, I would have imagined that I would have to call compare on one of the numbers (like calling it on an object in OOP). If 5 is Integer, which does implement Ord, then why do I call compare in the REPL itself? This is obviously a question related to a paradigm shift for me and I still didn't get it. Hopefully someone can explain.
The type defaulting here comes into play. The interpreter can derive that 5 and 7 need to be of the same type, and members of the Ord and Num typeclass. The default for a Num is Integer, and since Integer is an instance of Ord as well, we can thus use Integer.
The interpreter thus considers 5 and 7 to be Integers here in that case, and thus it can evaluate the function and obtain LT.
GHCi has some additional defaulting rules, described in the GHCi documentation.
Methods like compare are associated with types, not particular values. The compiler needs to be able to deduce the type in order to select the correct typeclass instance, but that doesn't require any special assistance.
The type of compare is
compare :: (Ord a) => a -> a -> Ordering
Thus any of its arguments (of type a) can be used to look up the Ord instance.
As you correctly assumed, in the compare 5 7 example, the types of 5 and 7 default to Integer. Thus a in the compare type is deduced to be Integer and the Ord Integer instance is selected.
This selection does not necessarily go through a function argument. Consider e.g.
read :: (Read a) => String -> a
Here it is the result type that drives instance selection, but the type checker is just fine with it:
> read "(2, 3)" :: (Int, Int)
(2,3)
(What would the OO equivalent be? "(2, 3)".read()?)
In fact, methods don't even have to be functions:
maxBound :: (Bounded a) => a
This is a polymorphic value, not a function:
> maxBound :: Int
9223372036854775807
Class instances are uniquely connected to types, so as long as the type checker has enough information to figure out what that type variable represents, everything works out. That is, in
someMethod :: (SomeClass foo) => ...
foo has to appear somewhere in the type signature ... so the type checker can resolve SomeClass foo from the way someMethod is used at any given point (at least in the absence of certain language extensions).
Warning: very beginner question.
I'm currently mired in the section on algebraic types in the Haskell book I'm reading, and I've come across the following example:
data Id a =
MkId a deriving (Eq, Show)
idInt :: Id Integer
idInt = MkId 10
idIdentity :: Id (a -> a)
idIdentity = MkId $ \x -> x
OK, hold on. I don't fully understand the idIdentity example. The explanation in the book is that:
This is a little odd. The type Id takes an argument and the data
constructor MkId takes an argument of the corresponding polymorphic
type. So, in order to have a value of type Id Integer, we need to
apply a -> Id a to an Integer value. This binds the a type variable to
Integer and applies away the (->) in the type constructor, giving us
Id Integer. We can also construct a MkId value that is an identity
function by binding the a to a polymorphic function in both the type
and the term level.
But wait. Why only fully polymorphic functions? My previous understanding was that a can be any type. But apparently constrained polymorphic type doesn't work: (Num a) => a -> a won't work here, and the GHC error suggests that only completely polymorphic types or "qualified types" (not sure what those are) are valid:
f :: (Num a) => a -> a
f = undefined
idConsPoly :: Id (Num a) => a -> a
idConsPoly = MkId undefined
Illegal polymorphic or qualified type: Num a => a -> a
Perhaps you intended to use ImpredicativeTypes
In the type signature for ‘idIdentity’:
idIdentity :: Id (Num a => a -> a)
EDIT: I'm a bonehead. I wrote the type signature below incorrectly, as pointed out by #chepner in his answer below. This also resolves my confusion in the next sentence below...
In retrospect, this behavior makes sense because I haven't defined a Num instance for Id. But then what explains me being able to apply a type like Integer in idInt :: Id Integer?
So in generality, I guess my question is: What specifically is the set of valid inputs to type constructors? Only fully polymorphic types? What are "qualified types" then? Etc...
You just have the type constructor in the wrong place. The following is fine:
idConsPoly :: Num a => Id (a -> a)
idConsPoly = MkId undefined
The type constructor Id here has kind * -> *, which means you can give it any value that has kind * (which includes all "ordinary" types) and returns a new value of kind *. In general, you are more concerned with arrow-kinded functions(?), of which type constructors are just one example.
TypeProd is a ternary type constructor whose first two arguments have kind * -> *:
-- Based on :*: from Control.Compose
newtype TypeProd f g a = Prod { unProd :: (f a, g a) }
Either Int is an expression whose value has kind * -> * but is not a type constructor, being the partial application of the type constructor Either to the nullary type constructor Int.
Also contributing to your confusion is that you've misinterpreted the error message from GHC. It means "Num a => a -> a is a polymorphic or qualified type, and therefore illegal (in this context)".
The last sentence that you quoted from the book is not very well worded, and maybe contributed to that misunderstanding. It's important to realize that in Id (a -> a) the argument a -> a is not a polymorphic type, but just an ordinary type that happens to mention a type variable. The thing which is polymorphic is idIdentity, which can have the type Id (a -> a) for any type a.
In standard Haskell polymorphism and qualification can only appear at the outermost level of the type in a type signature.
The type signature is almost correct
idConsPoly :: (Num a) => Id (a -> a)
Should be right, though i have no ghc on my phone to test this.
Also i think your question is quite broad, thus i deliberately answer only the concrete problem here.
As an example, consider the trivial function
f :: (Integral b) => a -> b
f x = 3 :: Int
GHC complains that it cannot deduce (b ~ Int). The definition matches the signature in the sense that it returns something that is Integral (namely an Int). Why would/should GHC force me to use a more specific type signature?
Thanks
Type variables in Haskell are universally quantified, so Integral b => b doesn't just mean some Integral type, it means any Integral type. In other words, the caller gets to pick which concrete types should be used. Therefore, it is obviously a type error for the function to always return an Int when the type signature says I should be able to choose any Integral type, e.g. Integer or Word64.
There are extensions which allow you to use existentially quantified type variables, but they are more cumbersome to work with, since they require a wrapper type (in order to store the type class dictionary). Most of the time, it is best to avoid them. But if you did want to use existential types, it would look something like this:
{-# LANGUAGE ExistentialQuantification #-}
data SomeIntegral = forall a. Integral a => SomeIntegral a
f :: a -> SomeIntegral
f x = SomeIntegral (3 :: Int)
Code using this function would then have to be polymorphic enough to work with any Integral type. We also have to pattern match using case instead of let to keep GHC's brain from exploding.
> case f True of SomeIntegral x -> toInteger x
3
> :t toInteger
toInteger :: Integral a => a -> Integer
In the above example, you can think of x as having the type exists b. Integral b => b, i.e. some unknown Integral type.
The most general type of your function is
f :: a -> Int
With a type annotation, you can only demand that you want a more specific type, for example
f :: Bool -> Int
but you cannot declare a less specific type.
The Haskell type system does not allow you to make promises that are not warranted by your code.
As others have said, in Haskell if a function returns a result of type x, that means that the caller gets to decide what the actual type is. Not the function itself. In other words, the function must be able to return any possible type matching the signature.
This is different to most OOP languages, where a signature like this would mean that the function gets to choose what it returns. Apparently this confuses a few people...
I'm new to Haskell and, in general, to functional programming, and I'm a bit uncomfortable with its syntax.
In the following code what does the => denote? And also (Num a, Ord a)?
loop :: (Num a, Ord a) => a -> (t -> t) -> t -> t
This is a typeclass constraint; (Num a, Ord a) => ... means that loop works with any type a that is an instance of the Num and Ord typeclasses, corresponding to numeric types and ordered types respectively. Basically, you can think of loop as having the type on the right hand side of the =>, except that a is required to be an instance of Num and Ord.
You can think of typeclasses as basically similar to OOP interfaces (but they're not the same thing!) — they encapsulate a set of definitions which any instance must support, and generic code can be written using these definitions. For instance, Num includes numeric operations like addition and multiplication, while Ord includes less than, greater than, and so on.
For more information on typeclasses, see this introduction from Learn You a Haskell.
=> separates two parts of a type signature:
On the left, typeclass constraints
On the right, the actual type
So you can think of (Num a, Ord a) => a -> (t -> t) -> t -> t as meaning "the type is a -> (t -> t) -> t -> t and also there must be a Num instance for a and an Ord instance for a".
For more on typeclasses see http://www.learnyouahaskell.com/types-and-typeclasses
One way to think about it is that Ord a and Num a are additional inputs to the function. They are a special kind of input though: dictionaries. When you use this function with a particular type a, there must also be dictionaries available for the Ord and Num operations on the type a as well.
Any function that makes use of a function with dictionary inputs must also have the same dictionary inputs.
foo :: (Num a, Ord a) => a -> t
foo x = loop x someFunc someT
However, you do not have to explicitly pass these dictionaries around. Haskell will take care of that for you, assuming there is a dictionary available. You can create a dictionary with a typeclass instance.
instance Num MyType with
x + y = ...
x - y = ...
...
This creates a dictionary for the Num operations on MyType, therefore MyType can be used anywhere that Num a is a required input (assuming it satisfies the other requirements, of course).
On the left hand side of the => you declare constraints for the types that are used on the right.
In the example you give, it means that a is constrained to being an instance of both the Ord type class and the Num type class.