I need to use this data structure data D = C Int Float and I need to compare it with an Int, for example a::D == 2.
How can I create an instance to define this kind of Eq ?
Thank you!
I would implement a projection:
getInt :: D -> Int
getInt (C i _) = i
and then compare with it:
getInt myD == 5
you can even include this into a record:
data D = C { getInt :: Int, getFloat :: Float }
if you like
You can't; == has signature a -> a -> Bool, so it can't be used like this.
Using the convertible package you can define
(~==) :: (Convertible a b, Eq b) => a -> b -> Bool
x ~== y = case safeConvert x of
Right x' -> x' == y
Left _ -> False
(==~) :: (Convertible b a, Eq a) => a -> b -> Bool
(==~) = flip (~==)
instance Convertible Int D where ...
-- or D Int depending on what you have in mind
Without depending on convertible, you could just define a conversion function either from Int to D (and write a == fromInt 2) or vice versa.
A less recommended route (for this specific case I think it's simply worse than the first solution) would be to define your own type class, e.g.
class Eq' a b where
(=~=) :: a -> b -> Bool
instance Eq a => Eq' a a where
x =~= y = x == y
instance Eq' D Int where ...
etc.
This can be done, but I doubt you will want to do this. As Alexey mentioned the type of (==) is Eq a=>a->a->Bool, so the only way to make this work would be to make 2 have type D. This may seem absurd at first, but in fact numbers can be made to have any type you want, as long as that type is an instance of Num
instance Num D where
fromInteger x = C x 1.0
There are still many things to work out, though....
First, you need to fully implement all the functions in Num, including (+), (*), abs, signum, fromInteger, and (negate | (-)).
Ugh!
Second, you have that extra Float to fill in in fromInteger. I chose the value 1.0 above, but that was arbitrary.
Third, you need to actually make D an instance of Eq also, to fill in the actual (==).
instance Eq D where
(C x _) == (C y _) = x == y
Note that this is also pretty arbitrary, as I needed to ignore the Float values to get (==) to do what you want it to.
Bottom line is, this would do what you want it to do, but at the cost of abusing the Num type, and the Eq type pretty badly.... The Num type should be reserved for things that you actually would think of as a number, and the Eq type should be reserved for a comparison of two full objects, each part included.
Related
In Haskell, when we talk type declaration.
I've seen both -> and =>.
As an example: I can make my own type declaration.
addMe :: Int -> Int -> Int
addMe x y = x + y
And it works just fine.
But if we take a look at :t sqrt we get:
sqrt :: Floating a => a -> a
At what point do we use => and when do we use ->?
When do we use "fat arrow" and when do we use "thin arrow"?
-> is for explicit functions. I.e. when f is something that can be written in an expression of the form f x, the signature must have one of these arrows in it†. Specifically, the type of x (the argument) must appear to the left of a -> arrow.
It's best to not think of => as a function arrow at all, at least at first‡. It's an implication arrow in the logical sense: if a is a type with the property Floating a, then it follows that the signature of sqrt is a -> a.
For your addMe example, which is a function with two arguments, the signature must always have the form x -> y -> z. Possibly there can also be a q => in front of that; that doesn't influence the function-ishness, but may have some saying in what particular types are allowed. Generally, such constraints are not needed if the types are already fixed and concrete. Like, you could in principle impose a constraint on Int:
addMe :: Num Int => Int -> Int -> Int
addMe x y = x + y
...but that doesn't really accomplish anything, because everybody knows that the particular type Int is an instance of the Num class. Where you need such constraints is when the type is not fixed but a type variable (i.e. lowercase), i.e. if the function is polymorphic. You can't just write
addMe' :: a -> a -> a
addMe' x y = x + y
because that signature would suggest the function works for any type a whatsoever, but it can't work for all types (how would you add, for example, two strings? ok perhaps not the best example, but how would you multiply two strings?)
Hence you need the constraint
addMe' :: Num a => a -> a -> a
addMe' x y = x + y
This means, you don't care what exact type a is, but you do require it to be a numerical type. Anybody can use the function with their own type MyNumType, but they need to ensure that Num MyNumType is fulfilled: then it follows that addMe' can have signature MyNumType -> MyNumType -> MyNumType.
The way to ensure this is to either use a standard type which you know to be numerical, for instance addMe' 5.9 3.7 :: Double would work, or give an instance declaration for your custom type and the Num class. Only do the latter if you're sure it's a good idea; usually the standard num types are all you'll need.
†Note that the arrow may not be visible in the signature: it's possible to have a type synonym for a function type, for example when type IntEndofunc = Int -> Int, then f :: IntEndofunc; f x = x+x is ok. But you can think of the typedef as essentially just a syntactic wrapper; it's still the same type and does have the arrow in it.
‡It so happens that logical implication and function application can be seen as two aspects of the same mathematical concept. Furthermore, GHC actually implements class constraints as function arguments, so-called dictionaries. But all this happens behind the scenes, so if anything they're implicit functions. In standard Haskell, you will never see the LHS of a => type as the type of some actual argument the function is applied to.
The "thin arrow" is used for function types (t1 -> t2 being the type of a function that takes a value of type t1 and produces a value of type t2).
The "fat arrow" is used for type constraints. It separates the list of type constraints on a polymorphic function from the rest of the type. So given Floating a => a -> a, we have the function type a -> a, the type of a function that can take arguments of any type a and produces a result of that same type, with the added constraint Floating a, meaning that the function can in fact only be used with types that implement the Floating type class.
the -> is the constructor of functions and the => is used to constraints, a sort of "interface" in Haskell called typeclass.
A little example:
sum :: Int -> Int -> Int
sum x y = x + y
that function only allows Int types, but if you want a huge int or a small int, you probably want Integer, and how to tell it to use both?
sum2 :: Integral a => a -> a -> a
sum2 x y = x + y
now if you try to do:
sum2 3 1.5
it will give you an error
also, you may want to know if two data are equals, you want:
equals :: Eq a => a -> a -> Bool
equals x y = x == y
now if you do:
3 == 4
that's ok
but if you create:
data T = A | B
equals A B
it will give to you:
error:
• No instance for (Eq T) arising from a use of ‘equals’
• In the expression: equals A B
In an equation for ‘it’: it = equals A B
if you want for that to work, you must just do:
data T = A | B deriving Eq
equals A B
False
I'm learning how to use typeclasses in Haskell.
Consider the following implementation of a typeclass T with a type constrained class function f.
class T t where
f :: (Eq u) => t -> u
data T_Impl = T_Impl_Bool Bool | T_Impl_Int Int | T_Impl_Float Float
instance T T_Impl where
f (T_Impl_Bool x) = x
f (T_Impl_Int x) = x
f (T_Impl_Float x) = x
When I load this into GHCI 7.10.2, I get the following error:
Couldn't match expected type ‘u’ with actual type ‘Float’
‘u’ is a rigid type variable bound by
the type signature for f :: Eq u => T_Impl -> u
at generics.hs:6:5
Relevant bindings include
f :: T_Impl -> u (bound at generics.hs:6:5)
In the expression: x
In an equation for ‘f’: f (T_Impl_Float x) = x
What am I doing/understanding wrong? It seems reasonable to me that one would want to specialize a typeclass in an instance by providing an accompaning data constructor and function implementation. The part
Couldn't match expected type 'u' with actual type 'Float'
is especially confusing. Why does u not match Float if u only has the constraint that it must qualify as an Eq type (Floats do that afaik)?
The signature
f :: (Eq u) => t -> u
means that the caller can pick t and u as wanted, with the only burden of ensuring that u is of class Eq (and t of class T -- in class methods there's an implicit T t constraint).
It does not mean that the implementation can choose any u.
So, the caller can use f in any of these ways: (with t in class T)
f :: t -> Bool
f :: t -> Char
f :: t -> Int
...
The compiler is complaining that your implementation is not general enough to cover all these cases.
Couldn't match expected type ‘u’ with actual type ‘Float’
means "You gave me a Float, but you must provide a value of the general type u (where u will be chosen by the caller)"
Chi has already pointed out why your code doesn't compile. But it's not even that typeclasses are the problem; indeed, your example has only one instance, so it might just as well be a normal function rather than a class.
Fundamentally, the problem is that you're trying to do something like
foobar :: Show x => Either Int Bool -> x
foobar (Left x) = x
foobar (Right x) = x
This won't work. It tries to make foobar return a different type depending on the value you feed it at run-time. But in Haskell, all types must be 100% determined at compile-time. So this cannot work.
There are several things you can do, however.
First of all, you can do this:
foo :: Either Int Bool -> String
foo (Left x) = show x
foo (Right x) = show x
In other words, rather than return something showable, actually show it. That means the result type is always String. It means that which version of show gets called will vary at run-time, but that's fine. Code paths can vary at run-time, it's types which cannot.
Another thing you can do is this:
toInt :: Either Int Bool -> Maybe Int
toInt (Left x) = Just x
toInt (Right x) = Nothing
toBool :: Either Int Bool -> Maybe Bool
toBool (Left x) = Nothing
toBool (Right x) = Just x
Again, that works perfectly fine.
There are other things you can do; without knowing why you want this, it's difficult to suggest others.
As a side note, you want to stop thinking about this like it's object oriented programming. It isn't. It requires a new way of thinking. In particular, don't reach for a typeclass unless you really need one. (I realise this particular example may just be a learning exercise to learn about typeclasses of course...)
It's possible to do this:
class Eq u => T t u | t -> u where
f :: t -> u
You need FlexibleContextx+FunctionalDepencencies and MultiParamTypeClasses+FlexibleInstances on call-site. Or to eliminate class and to use data types instead like Gabriel shows here
-- | Convert a 'Maybe a' to an equivalent 'Either () a'. Should be inverse
-- to 'eitherUnitToMaybe'.
maybeToEitherUnit :: Maybe a -> Either () a
maybeToEitherUnit a = error "Not yet implemented: maybeToEitherUnit"
-- | Convert a 'Either () a' to an equivalent 'Maybe a'. Should be inverse
-- to 'maybeToEitherUnit'.
eitherUnitToMaybe :: Either () a -> Maybe a
eitherUnitToMaybe = error "Not yet implemented: eitherUnitToMaybe"
-- | Convert a pair of a 'Bool' and an 'a' to 'Either a a'. Should be inverse
-- to 'eitherToPairWithBool'.
pairWithBoolToEither :: (Bool,a) -> Either a a
pairWithBoolToEither = undefined -- What should I do here?
-- | Convert an 'Either a a' to a pair of a 'Bool' and an 'a'. Should be inverse
-- to 'pairWithBoolToEither'.
eitherToPairWithBool :: Either a a -> (Bool,a)
eitherToPairWithBool = undefined -- What should I do here?
-- | Convert a function from 'Bool' to 'a' to a pair of 'a's. Should be inverse
-- to 'pairToFunctionFromBool'.
functionFromBoolToPair :: (Bool -> a) -> (a,a)
functionFromBoolToPair = error "Not yet implemented: functionFromBoolToPair"
-- | Convert a pair of 'a's to a function from 'Bool' to 'a'. Should be inverse
-- to 'functionFromBoolToPair'.
pairToFunctionFromBool :: (a,a) -> (Bool -> a)
pairToFunctionFromBool = error "Not yet implemented: pairToFunctionFromBool"
I don't really know what to do. I know what maybe is, but I think I have a problem with either, because Either a a makes no sense in my mind. Either a b would be okay. This is either a or b but Either a a is a?!
I don't have any idea in general how to write these functions.
Given that I think this is homework, I'll not answer, but give important hints:
If you look for the definitions on hoogle (http://www.haskell.org/hoogle/)
you find
data Bool = True | False
data Either a b = Left a | Right b
This means that Bool can only be True or False, but that Either a b can be Left a or Right b.
which means your functions should look like
pairWithBoolToEither :: (Bool,a) -> Either a a
pairWithBoolToEither (True,a) = ....
pairWithBoolToEither (False,a) = ....
and
eitherToPairWithBool :: Either a a -> (Bool,a)
eitherToPairWithBool (Left a) = ....
eitherToPairWithBool (Right a) = ....
Comparing with Maybe
Maybe a is given by
data Maybe a = Just a | Nothing
so something of type Maybe Int could be Just 7 or Nothing.
Similarly, something of type Either Int Char could be Left 5 or Right 'c'.
Something of type Either Int Int could be Left 7 or Right 4.
So something with type Either Int Char is either an Int or a Char, but something of type Either Int Int is either an Int or an Int. You don't get to choose anything other than Int, but you'll know whether it was a Left or a Right.
Why you've been asked this/thinking behind it
If you have something of type Either a a, then the data (eg 5 in Left 5) is always of type a, and you've just tagged it with Left or Right. If you have something of type (Bool,a) the a-data (eg 5 in (True,5)) is always the same type, and you've paired it with False or True.
The maths word for two things which perhaps look different but actually have the same content is "isomorphic". Your instructor has asked you to write a pair of functions which show this isomorphism. Your answer will go down better if pairWithBoolToEither . eitherToPairWithBool and eitherToPairWithBool . pairWithBoolToEither do what id does, i.e. don't change anything. In fact, I've just spotted the comments in your question, where it says they should be inverses. In your write-up, you should show this by doing tests in ghci like
ghci> eitherToPairWithBool . pairWithBoolToEither $ (True,'h')
(True,'h')
and the other way round.
(In case you haven't seen it, $ is defined by f $ x = f x but $ has really low precedence (infixr 0 $), so f . g $ x is (f . g) $ x which is just (f . g) x and . is function composition, so (f.g) x = f (g x). That was a lot of explanation to save one pair of brackets!)
Functions that take or return functions
This can be a bit mind blowing at first when you're not used to it.
functionFromBoolToPair :: (Bool -> a) -> (a,a)
The only thing you can pattern match a function with is just a variable like f, so we'll need to do something like
functionFromBoolToPair f = ...
but what can we do with that f? Well, the easiest thing to do with a function you're given is to apply it to a value. What value(s) can we use f on? Well f :: (Bool -> a) so it takes a Bool and gives you an a, so we can either do f True or f False, and they'll give us two (probably different) values of type a. Now that's handy, because we needed to a values, didn't we?
Next have a look at
pairToFunctionFromBool :: (a,a) -> (Bool -> a)
The pattern match we can do for the type (a,a) is something like (x,y) so we'll need
pairToFunctionFromBool (x,y) = ....
but how can we return a function (Bool -> a) on the right hand side?
There are two ways I think you'll find easiest. One is to notice that since -> is right associative anyway, the type (a,a) -> (Bool -> a) is the same as (a,a) -> Bool -> a so we can actually move the arguments for the function we want to return to before the = sign, like this:
pairToFunctionFromBool (x,y) True = ....
pairToFunctionFromBool (x,y) False = ....
Another way, which feels perhaps a little easier, would to make a let or where clause to define a function called something like f, where f :: Bool -> a< a bit like:
pairToFunctionFromBool (x,y) = f where
f True = ....
f False = ....
Have fun. Mess around.
Perhaps it's useful to note that Either a b is also called the coproduct, or sum, of the types a and b. Indeed it is now common to use
type (+) = Either
You can then write Either a b as a + b.
eitherToPairWithBool :: (a+a) -> (Bool,a)
Now common sense would dictate that we rewrite a + a as something like 2 ⋅ a. Believe it or not, that is exactly the meaning of the tuple type you're transforming to!
To explain: algebraic data types can roughly be seen as "counting1 the number of possible constructions". So
data Bool = True | False
has two constructors. So sort of (this is not valid Haskell!)
type 2 = Bool
Tuples allow all the combinations of constructors from each argument. So for instance in (Bool, Bool), we have the values
(False,False)
(False,True )
(True, False)
(True, True )
You've guessed it: tuples are also called products. So the type (Bool, a) is basically 2 ⋅ a: for every value x :: a, we can create both the (False, x) tuple and the (True, x) tuple, alltogether twice as many as there are x values.
Much the same thing for Either a a: we always have both Left x and Right x as a possible value.
All your functions with "arithmetic types":
type OnePlus = Maybe
maybeToEitherUnit :: OnePlus a -> () + a
eitherUnitToMaybe :: () + a -> OnePlus a
pairWithBoolToEither :: 2 ⋅ a -> a + a
eitherToPairWithBool :: a + a -> 2 ⋅ a
functionFromBoolToPair :: a² -> a⋅a
pairToFunctionFromBool :: a⋅a -> a²
1For pretty much any interesting type there are actually infinitely many possible values, still this kind of naïve arithmetic gets you surprisingly far.
Either a a makes no sense in my mind.
Yes it does. Try to figure out the difference between type a and Either a a. Either is a disjoint union. Once you understand the difference between a and Either a a, your homework should be easy in conjunction with AndrewC's answer.
Note that Either a b means quite literally that a value of such a type can be either an a, or an a. It sounds like you have actually grasped this concept, but the piece you're missing is that the Either type differentiates between values constructed with Left and those constructed with Right.
For the first part, the idea is that Maybe is either Just a thing or Nothing -- Nothing corresponds to () because both are "in essence" data types with only one possible value.
The idea behind converting (Bool, a) pairs to Either a a pairs might seem a little trickier, but just think about the correspondence between True and False and Left and Right.
As for converting functions of type (Bool -> a) to (a, a) pairs, here's a hint: Consider the fact that Bool can only have two types, and write down what that initial function argument might look like.
Hopefully those hints help you to get started.
I am currently reading this, but if I am honest I am struggling to see what
class Eq a where
(==) :: a -> a -> Bool
achieves, which
instance Eq Integer where
x == y = x `integerEq` y
doesnt achieve. I understand the second code defines what the result of performing equality on two Integer types should be. What is the purpose of the first then??
The class declaration says "I'm going to define a bunch of functions now which will work for several different types". The instance declaration says "this is how these functions work for this type".
In your specific example, class Eq says that "Eq means any type that has a function named ==", whereas the instance Eq Integer says "this is how == works for an Integer".
The first defines what operations must be provided for a type to be comparable for equality. You can then use that to write functions that operate on any type that is comparable for equality, not just integers.
allSame :: Eq a => [a] -> Bool
allSame [] = []
allSame (x:xs) = foldr True (==x) xs
This function works for integers because instances for Eq Integer exists. It also works for strings ([Char]) because an instance for Eq Char exists, and an instance for lists of types that have instances of Eq also exists (instance Eq a => Eq [a]).
There is one class and many instances for different types. That's why the class specifies the required signature (interface; classes can also specify default implementations, but that's beside the point), and instance the body (implementation). You then use class name as a constraint that means "any type a that implements Eq operations, i.e. have an instance in Eq".
Read Learn you a Haskell or Real World Haskell, they're better than the haskell.org tutorial.
Let's say you want to implement a generic algorithm or data structure, "generic" meaning polymorphic: it should work for any data type. For example, let's say you want to write a function that determines whether three input values are equal.
Taking a specific (monomorphic) case, you can do this for integers:
eq3 :: Int -> Int -> Int -> Bool
eq3 x y z = x == y && y == z
We'd expect the above definition to work for other types as well, of course, but if we simply tell the compiler that the function should apply to any type:
eq3 :: a -> a -> a -> Bool
eq3 x y z = x == y && y == z
... the compiler complains that the == function doesn't apply to our generic a:
<interactive>:12:49:
No instance for (Eq a)
arising from a use of `=='
In the first argument of `(&&)', namely `x == y'
In the expression: x == y && y == z
In an equation for `eq3': eq3 x y z = x == y && y == z
We have to tell the compiler that our type a is an instance of the Eq type class, which you already noticed is where the == function is declared. See the difference here:
eq3 :: Eq a => a -> a -> a -> Bool
eq3 x y z = x == y && y == z
Now we have a function that can operate uniformly on any type a belonging to the Eq type class.
I need a function which works like:
some :: (Int, Maybe Int) -> Int
some a b
| b == Nothing = 0
| otherwise = a + b
Use cases:
some (2,Just 1)
some (3,Nothing)
map some [(2, Just 1), (3,Nothing)]
But my code raise the error:
The equation(s) for `some' have two arguments,
but its type `(Int, Maybe Int) -> Int' has only one
I don't understand it.
Thanks in advance.
When you write
foo x y = ...
That is notation for a curried function, with a type like:
foo :: a -> b -> c
You have declared your function to expect a tuple, so you must write it:
some :: (Int, Maybe Int) -> Int
some (x, y) = ...
But Haskell convention is usually to take arguments in the former curried form. Seeing funcitons take tuples as arguments is very rare.
For the other part of your question, you probably want to express it with pattern matching. You could say:
foo :: Maybe Int -> Int
foo Nothing = 0
foo (Just x) = x + 1
Generalizing that to the OP's question is left as an exercise for the reader.
Your error doesn't come from a misunderstanding of Maybe: The type signature of some indicates that it takes a pair (Int, Maybe Int), while in your definition you provide it two arguments. The definition should thus begin with some (a,b) to match the type signature.
One way to fix the problem (which is also a bit more idiomatic and uses pattern matching) is:
some :: (Int, Maybe Int) -> Int
some (a, Nothing) = a
some (a, Just b) = a + b
It's also worth noting that unless you have a really good reason for using a tuple as input, you should probably not do so. If your signature were instead some :: Int -> Maybe Int -> Int, you'd have a function of two arguments, which can be curried. Then you'd write something like
some :: Int -> Maybe Int -> Int
some a Nothing = a
some a (Just b) = a + b
Also, you might want to add the following immediate generalization: All Num types are additive, so you might aswell do
some :: (Num n) => n -> Maybe n -> n
some a Nothing = a
some a (Just b) = a + b
(I've violated the common practice of using a, b, c... for type variables so as not to confuse the OP since he binds a and b to the arguments of some).