I'm trying to define a Complex datatype, and I want the constructors to be able to take any number as instance, so I would like to use a generic type as long as it does implement a Num instance
I'm usind GADTs in order to do so since to my understanding the DataTypeContexts language extension was a "misfeature" even if I think that would have been useful in this case...
In any case this is my code:
data Complex where
Real :: ( Num num, Show num ) => num -> Complex
Imaginary :: ( Num num, Show num ) => num -> Complex
Complex :: ( Num num, Show num ) => num -> num -> Complex
real :: ( Num num, Show num ) => Complex -> num
real (Real r) = r
real (Imaginary _i ) = 0
real (Complex r _i ) = r
here the real implementation gives the following error:
Couldn't match expected type ‘num’ with actual type ‘num1’
‘num1’ is a rigid type variable bound by
a pattern with constructor:
Real :: forall num. (Num num, Show num) => num -> Complex,
in an equation for ‘real’
at <...>/Complex.hs:29:7-12
‘num’ is a rigid type variable bound by
the type signature for:
real :: forall num. (Num num, Show num) => Complex -> num
at <...>/Complex.hs:28:1-47
• In the expression: r
In an equation for ‘real’: real (Real r) = r
• Relevant bindings include
r :: num1
(bound at <...>/Complex.hs:29:12)
real :: Complex -> num
(bound at <...>/Complex.hs:29:1)
which to my understanding is due to the return type do be interpreted as different...
so I tried removing the type definition and letting ghc do his magic with the type but turns out the type signature was the same...
can anyone please explain to me what is wrong here?
Problem is, these definitions allow you to choose different types when (1) constructing a Complex value and when (2) applying the real function. These two situations are not connected to each other in any way, so there is nothing to force the type to be the same between them. For example:
c :: Complex
c = Real (42 :: Int)
d :: Double
d = real c
The definition of d requires the real function to return a Double, but there is no Double wrapped inside of c, there is only Int.
As for solutions, there are two possible ones: (1) establish a connection between these two points, forcing the type to be the same, and (2) allow the type inside to be converted to any other numeric type.
To establish a type-level connection between two points, we need to use a type that is present at both points. What type would that be? Quite obviously, that's the type of c. So we need to make the type of c somehow convey what's wrapped inside it:
data Complex num = Real num | Imaginary num | Complex num num
real :: Complex num -> num
real = ...
-- Usage:
c :: Complex Int
c = Real 42
d :: Int
d = real c
Note that I don't actually need GADTs for this.
To allow type conversion, you'll need to require another type class for the num type. The class Num has a way to convert from any integral type, but there is no way to convert to any such type, because it doesn't make sense: 3.1415 can't be meaningfully converted to an integral type.
So you'll have to come up with your own way to convert, and implement it for all allowed types too:
class Convert a where
toNum :: Num n => a -> n
data Complex where
Real :: ( Num num, Show num, Convert num ) => num -> Complex
...
real :: Num num => Complex -> num
real (Real r) = toNum r
...
Just to be clear, I consider the second option quite insane. I only provided it for illustration. Don't do it. Go with option 1.
Related
I am going through the Haskell Book (http://haskellbook.com/) and got stuck at the following exercise:
f :: Float
f = 1.0
-- Question: Can you replace `f`'s signature by `f :: Num a => a`
At first, I thought the answer would be Yes. Float provides an instance for Num, so substituting a Num a => a by a Float value should be fine (I am thinking co-variance here).
This does not compile however:
Could not deduce (Fractional a) arising from the literal ‘1.0’
from the context: Num a
bound by the type signature for:
f :: forall a. Num a => a
at ...
Possible fix:
add (Fractional a) to the context of
the type signature for:
f :: forall a. Num a => a
• In the expression: 1.0
In an equation for ‘f’: f = 1.0
If I do this however, no problem:
f :: Fractional a => a
f = 1.0
Why can't I use a less specific constraint here like Num a => a?
UPD:
Actually, you can sum this up to:
1.0::(Num a => a)
vs
1.0::(Fractional a => a)
Why is the second working but not the first one? I thought Fractional was a subset of Num (meaning Fractional is compatible with Num)
UPD 2:
Thanks for your comments, but I am still confused. Why this works:
f :: Num a => a -> a
f a = a
f 1.0
while this not:
f :: Num a => a
f = 1.0
UPD 3:
I have just noticed something:
f :: Num a => a
f = (1::Int)
does not work either.
UPD 4
I have been reading all the answers/comments and as far as I understand:
f :: Num a => a
is the Scala equivalent of
def f[A: Num]: A
which would explain why many mentioned that a is defined by the caller. The only reason why we can write this:
f :: Num a => a
f = 1
is because 1 is typed as a Num a => a. Could someone please confirm this assumption? In any case, thank you all for your help.
If I have f :: Num a => a this means that I can use f wherever I need any numeric type. So, all of f :: Int, f :: Double must type check.
In your case, we can't have 1.0 :: Int for the same reason we can't have 543.24 :: Int, that is, Int does not represent fractional values. However, 1.0 does fit any fractional type (as 543.24 does).
Fractional indeed can be thought as a subset of Num. But if I have a value inside all the fractionals f :: forall a . Fractional a => a, I don't necessarily have a value in all the numeric types f :: forall a . Num a => a.
Note that, in a sense, the constraints are on the left side of =>, which makes them behave contra-variantly. I.e. cars are a subset of vehichles, but I can't conclude that a wheel that can be used in any car will be able to be used with any vehicle. Rather, the opposite: a wheel that can be used in any vehicle will be able to be used with any car.
So, you can roughly regard f :: forall a . Num a => a (values fitting in any numeric type, like 3 and 5) as a subtype of f :: forall a . Fractional a => a (values fitting in any fractional type, like 3,5, and 32.43).
Let's start with the monomorphic case:
f :: Float
f = 1.0
Here, you've said that f is a Float; not an Int, not a Double, not any other type. 1.0, on the other hand, is a polymorphic constant; it has type Fractional a => a, and so can be used to provide a value of any type that has a Fractional instance. You can restrict it to Float, or Double, etc. Since f has to be a Float, that's what 1.0 is restricted to.
If you try to change the signature
f :: Num a => a
f = 1.0
now you have a problem. You've now promised that f can be used to provide a value of any type that has a Num instance. That includes Int and Integer, but neither of those types has a Fractional instance. So, the compiler refuses to let you do this. 1.0 simply can't produce an Int should the need arise.
Likewise,
1.0::(Num a => a)
is a lie, because this isn't a restriction; it's an attempt at expanding the set of types that 1.0 can produce. Something with type Num a => a should be able to give me an Int, but 1.0 cannot do that.
1.0::(Fractional a => a)
works because you are just restating something that is already true. It's neither restricting 1.0 to a smaller set of types or trying to expand it.
Now we get something a little more interesting, because you are specifying a function type, not just a value type.
f :: Num a => a -> a
f a = a
This just says that f can take as its argument any value that is no more polymorphic than Num a => a. a can be any type that implements Num, or it can be a polymorphic value that is a subset of the types that represent Num.
You chose
f 1.0
which means a gets unified with Fractional a => a. Type inference then decides that the return type is also Fractional a => a, and returning the same value you passed in is allowed.
We've already covered why this
f :: Num a => a
f = 1.0
isn't allowed above, and
f :: Num a => a
f = (1::Int)
fails for the same reason. Int is simply too specific; it is not the same as Num a => a.
For example, (+) :: Num a => a -> a -> a requires two arguments of the same type. So, I might try to write
1 :: Integer + f
or
1 :: Float + f
. In both cases, I need f to be able to provide a value with the same type as the other argument, to satisfy the type of (+): if I want an Integer, I should be able to get an Integer, and if I want a Float, I should be able to get a Float. But if you could specify a value with something less specific than Num a => a, you wouldn't be able to keep that promise. 1.0 :: Fractional a => a can't provide an Integer, and 1 :: Int can't provide anything except an Int, not even an Integer.
Think of a polymorphic type as a function from a concrete type to a value; you can literally do this if you enable the TypeApplications extension.
Prelude> :set -XTypeApplications
Prelude> :{
Prelude| f :: Num a => a
Prelude| f = 1
Prelude| :}
Prelude> :t f
f :: Num a => a
Prelude> :t f #Int
f #Int :: Int
Prelude> f #Int
1
Prelude> :t f #Float
f #Float :: Float
Prelude> f #Float
1.0
Prelude> :t f #Rational
f #Rational :: Rational
Prelude> f #Rational
1 % 1
The reason these all work is because you promised that you could pass any type with a Num instance to f, and it could return a value of that type. But if you had been allowed to say f = 1.0, there is no way that f #Int could, in fact, return an Int, because 1.0 simply is not capable of producing an Int.
When values with polymorphic types like Num a => a are involved, there are two sides. The source of the value, and the use of the value. One side gets the flexibility to choose any specific type compatible with the polymorphic type (like using Float for Num a => a). The other side is restricted to using code that will work regardless of which specific type is involved - it can only make use of features that will work for every type compatible with the polymorphic type.
There is no free lunch; both sides cannot have the same freedom to pick any compatible type they like.
This is just as true for object-oriented subclass polymorphism, however OO polymorphism rules give the flexibility to the source side, and put all the restrictions on the use side (except for generics, which works like parametric polymorphism), while Haskell's parametric polymorphism gives the flexibility to the use side, and puts all the restrictions on the source side.
For example in an OO language with a general number class Num, and Int and Double as subclasses of this, then a method returning something of type Num would work the way you're expecting. You can return 1.0 :: Double, but the caller can't use any methods on the value that are provided specifically by Double (like say one that splits off the fractional part) because the caller must be programmed to work the same whether you return an Int or a Double (or even any brand new subclass of Num that is private to your code, that the caller cannot possibly know about).
Polymorphism in Haskell is based on type parameters rather than subtyping, which switches things around. The place in the code where f :: Num a => a is used has the freedom to demand any particular choice for a (subject to the Num constraint), and the code for f that is the source of the value must be programmed to work regardless of the use-site's choice. The use-site is even free to demand values of a type that is private to the code using f, that the implementer of f cannot possibly know about. (I can literally open up a new file, make any bizarre new type I like and give it an instance for Num, and any of the standard library functions written years ago that are polymorphic in Num will work with my type)
So this works:
f :: Float
f = 1.0
Because there are no type variables, so both source and use-site simply have to treat this as a Float. But this does not:
f :: Num a => a
f = 1.0
Because the place where f is used can demand any valid choice for a, and this code must be able to work for that choice. (It would not work when Int is chosen, for example, so the compiler must reject this definition of f). But this does work:
f :: Fractional a => a
f = 1.0
Because now the use site is only free to demand any type that is in Fractional, which excludes the ones (like Int) that floating point literals like 1.0 cannot support.
Note that this is exactly how "generics" work in object oriented languages, so if you're familiar with any language supporting generics, just treat Haskell types the way you do generic ones in OO languages.
One further thing that may be confusing you is that in this:
f :: Float
f = 1.0
The literal 1.0 isn't actually definitively a Float. Haskell literals are much more flexible than those of most other languages. Whereas e.g. Java says that 1.0 is definitely a value of type double (with some automatic conversion rules if you use a double where certain other types are expected), in Haskell that 1.0 is actually itself a thing with a polymorphic type. 1.0 has type Fractional a => a.
So the reason the f :: Fractional a => a definition worked is obvious, it's actually the f :: Float definition that needs some explanation. It's making use of exactly the rules I described above in the first section of my post. Your code f = 1.0 is a use-site of the value represented by 1.0, so it can demand any type it likes (subject to Fractional). In particular, it can demand that the literal 1.0 supply a value of type Float.
This again reinforces why the f :: Num a => a definition can't work. Here the type for f is promising to f's callers that they can demand any type they like (subject to Num). But it's going to fulfill that demand by just passing the demand down the chain to the literal 1.0, which has the most general type of Fractional a => a. So if the use-site of f demands a type that is in Num but outside Fractional, f would then try to demand that 1.0 supply that same non-Fractional type, which it can't.
Names do not have types. Values have types, and the type is an intrinsic part of the value. 1 :: Int and 1 :: Integer are two different values. Haskell does not implicitly convert values between types, though it can be trivial to define functions that take values of one type and return values of another. (For example, f :: Int -> Integer with f x = x will "convert" its Int arugment to an Integer.
A declaration like
f :: Num a => a
does not say that f has type Num a => a, it says that you can assign values of type Num a => a to f.
You can think of a polymorphic value like 1 :: Num a => a being all 1-like values in every type with a Num instance, including 1 :: Int, 1 :: Integer, 1 :: Rational, etc.
An assignment like f = 1 succeeds because the literal 1 has the expected type Num a => a.
An assignment like f = 1.0 fails because the literal 1.0 has a different type, Fractional a => a, and that type is too specific. It does not include all 1-like values that Num a => a may be called on to produce.
Suppose you declared g :: Fractional a => a. You can say g = 1.0, because the types match. You cannot say g = (1.0 :: Float), because the types do not match; Float has a Fractional instance, but it is just one of a possibly infinite set of types that could have Fractional instances.
You can say g = 1, because Fractional a => a is more specific than Num a => a, and has Fractional has Num as its superclass. The assignment "selects" the subset of 1 :: Num a => a that overlaps with (and for all intents and purposes is) 1 :: Fractional a => a and assigns that to g. Put another way, just 1 :: Num a => a can produce a value for any single type that has a Num instance, it can produce a value for any subset of types implied by a subclass of Num.
I am puzzled by this as well.
What gets stuck in my mind is something like:
// in pseudo-typescript
type Num = Int | Float;
let f: Num;
f = (1.0 as Float); // why this doesn't work
Fact is, Num a => a is just not a simple sum of numeric types.
It represents something that can morph into various kind of numeric types.
Thanks to chepner's explanation, now I can persuade myself like this:
if I have a Num a => a then I can get a Int from it, I can also get a Float from it, as well as a Double....
If I am able to install 1.1 into a Num a => a, then there is no way I can safely derive a Int from 1.1.
The expression 1 is able to bind to Num a => a is due to the fact that 1 itself is a polymorphic constant with type signature Num a => a.
So, I'm trying to write my own replacement for Prelude, and I have (^) implemented as such:
{-# LANGUAGE RebindableSyntax #-}
class Semigroup s where
infixl 7 *
(*) :: s -> s -> s
class (Semigroup m) => Monoid m where
one :: m
class (Ring a) => Numeric a where
fromIntegral :: (Integral i) => i -> a
fromFloating :: (Floating f) => f -> a
class (EuclideanDomain i, Numeric i, Enum i, Ord i) => Integral i where
toInteger :: i -> Integer
quot :: i -> i -> i
quot a b = let (q,r) = (quotRem a b) in q
rem :: i -> i -> i
rem a b = let (q,r) = (quotRem a b) in r
quotRem :: i -> i -> (i, i)
quotRem a b = let q = quot a b; r = rem a b in (q, r)
-- . . .
infixr 8 ^
(^) :: (Monoid m, Integral i) => m -> i -> m
(^) x i
| i == 0 = one
| True = let (d, m) = (divMod i 2)
rec = (x*x) ^ d in
if m == one then x*rec else rec
(Note that the Integral used here is one I defined, not the one in Prelude, although it is similar. Also, one is a polymorphic constant that's the identity under the monoidal operation.)
Numeric types are monoids, so I can try to do, say 2^3, but then the typechecker gives me:
*AlgebraicPrelude> 2^3
<interactive>:16:1: error:
* Could not deduce (Integral i0) arising from a use of `^'
from the context: Numeric m
bound by the inferred type of it :: Numeric m => m
at <interactive>:16:1-3
The type variable `i0' is ambiguous
These potential instances exist:
instance Integral Integer -- Defined at Numbers.hs:190:10
instance Integral Int -- Defined at Numbers.hs:207:10
* In the expression: 2 ^ 3
In an equation for `it': it = 2 ^ 3
<interactive>:16:3: error:
* Could not deduce (Numeric i0) arising from the literal `3'
from the context: Numeric m
bound by the inferred type of it :: Numeric m => m
at <interactive>:16:1-3
The type variable `i0' is ambiguous
These potential instances exist:
instance Numeric Integer -- Defined at Numbers.hs:294:10
instance Numeric Complex -- Defined at Numbers.hs:110:10
instance Numeric Rational -- Defined at Numbers.hs:306:10
...plus four others
(use -fprint-potential-instances to see them all)
* In the second argument of `(^)', namely `3'
In the expression: 2 ^ 3
In an equation for `it': it = 2 ^ 3
I get that this arises because Int and Integer are both Integral types, but then why is it that in normal Prelude I can do this just fine? :
Prelude> :t (2^)
(2^) :: (Num a, Integral b) => b -> a
Prelude> :t 3
3 :: Num p => p
Prelude> 2^3
8
Even though the signatures for partial application in mine look identical?
*AlgebraicPrelude> :t (2^)
(2^) :: (Numeric m, Integral i) => i -> m
*AlgebraicPrelude> :t 3
3 :: Numeric a => a
How would I make it so that 2^3 would in fact work, and thus give 8?
A Hindley-Milner type system doesn't really like having to default anything. In such a system, you want types to be either properly fixed (rigid, skolem) or properly polymorphic, but the concept of “this is, like, an integer... but if you prefer, I can also cast it to something else” as many other languages have doesn't really work out.
Consequently, Haskell sucks at defaulting. It doesn't have first-class support for that, only a pretty hacky ad-hoc, hard-coded mechanism which mainly deals with built-in number types, but fails at anything more involved.
You therefore should try to not rely on defaulting. My opinion is that the standard signature for ^ is unreasonable; a better signature would be
(^) :: Num a => a -> Int -> a
The Int is probably controversial – of course Integer would be safer in a sense; however, an exponent too big to fit in Int generally means the results will be totally off the scale anyway and couldn't feasibly be calculated by iterated multiplication; so this kind of expresses the intend pretty well. And it gives best performance for the extremely common situation where you just write x^2 or similar, which is something where you very definitely don't want to have to put an extra signature in the exponent.
In the rather fewer cases where you have a concrete e.g. Integer number and want to use it in the exponent, you can always shove in an explicit fromIntegral. That's not nice, but rather less of an inconvenience.
As a general rule, I try to avoid† any function-arguments that are more polymorphic than the results. Haskell's polymorphism works best “backwards”, i.e. the opposite way as in dynamic language: the caller requests what type the result should be, and the compiler figures out from this what the arguments should be. This works pretty much always, because as soon as the result is somehow used in the main program, the types in the whole computation have to be linked to a tree structure.
OTOH, inferring the type of the result is often problematic: arguments may be optional, may themselves be linked only to the result, or given as polymorphic constants like Haskell number literals. So, if i doesn't turn up in the result of ^, avoid letting in occur in the arguments either.
†“Avoid” doesn't mean I don't ever write them, I just don't do so unless there's a good reason.
I tested the numeric coercion by using GHCI:
>> let c = 1 :: Integer
>> 1 / 2
0.5
>> c / 2
<interactive>:15:1: error:
• No instance for (Fractional Integer) arising from a use of ‘/’
• In the expression: c / 2
In an equation for ‘it’: it = c / 2
>> :t (/)
(/) :: Fractional a => a -> a -> a -- (/) needs Fractional type
>> (fromInteger c) / 2
0.5
>>:t fromInteger
fromInteger :: Num a => Integer -> a -- Just convert the Integer to Num not to Fractional
I can use fromInteger function to convert a Integer type to Num (fromInteger has the type fromInteger :: Num a => Integer -> a), but I cannot understand that how can the type Num be converted to Fractional implicitly?
I know that if an instance has type Fractional it must have type Num (class Num a => Fractional a where), but does it necessary that if an instance has type Num it can be used as an instance with Fractional type?
#mnoronha Thanks for your detailed reply. There is only one question confuse me. I know the reason that type a cannot be used in function (/) is that type a is with type Integer which is not an instance of type class Fractional (the function (/) requires that the type of arguments must be instance of Fractional). What I don't understand is that even by calling fromInteger to convert the type integer to atype which be an instance of Num, it does not mean a type be an instance of Fractional (because Fractional type class is more constrained than Num type class, so a type may not implement some functions required by Fractional type class). If a type does not fully fit the condition Fractional type class requires, how can it be use in the function (/) which asks the arguments type be instance of Fractional. Sorry for not native speaker and really thanks for your patience!
I tested that if a type only fits the parent type class, it cannot be used in a function which requires more constrained type class.
{-# LANGUAGE OverloadedStrings #-}
module Main where
class ParentAPI a where
printPar :: int -> a -> String
class (ParentAPI a) => SubAPI a where
printSub :: a -> String
data ParentDT = ParentDT Int
instance ParentAPI ParentDT where
printPar i p = "par"
testF :: (SubAPI a) => a -> String
testF a = printSub a
main = do
let m = testF $ ParentDT 10000
return ()
====
test-typeclass.hs:19:11: error:
• No instance for (SubAPI ParentDT) arising from a use of ‘testF’
• In the expression: testF $ ParentDT 10000
In an equation for ‘m’: m = testF $ ParentDT 10000
In the expression:
do { let m = testF $ ParentDT 10000;
return () }
I have found a doc explaining the numeric overloading ambiguity very clearly and may help others with the same confusion.
https://www.haskell.org/tutorial/numbers.html
First, note that both Fractional and Num are not types, but type classes. You can read more about them in the documentation or elsewhere, but the basic idea is that they define behaviors for types. Num is the most inclusive numeric typeclass, defining behaviors functions like (+), negate, which are common to pretty much all "numeric types." Fractional is a more constrained type class that describes "fractional numbers, supporting real division."
If we look at the type class definition for Fractional, we see that it is actually defined as a subclass of Num. That is, for a type a to be an have an instance Fractional, it must first be a member of the typeclass Num:
class Num a => Fractional a where
Let's consider some type that is constrained by Fractional. We know it implements the basic behaviors common to all members of Num. However, we can't expect it to implement behaviors from other type classes unless multiple constraints are specified (ex. (Num a, Ord a) => a. Take, for example, the function div :: Integral a => a -> a -> a (integral division). If we try to apply the function with an argument that is constrained by the typeclass Fractional (ex. 1.2 :: Fractional t => t), we encounter an error. Type classes restrict the sort of values a function deals with, allowing us to write more specific and useful functions for types that share behaviors.
Now let's look at the more general typeclass, Num. If we have a type variable a that is only constrained by Num a => a, we know that it will implement the (few) basic behaviors included in the Num type class definition, but we'd need more context to know more. What does this mean practically? We know from our Fractional class declaration that functions defined in the Fractional type class are applied to Num types. However, these Num types are a subset of all possible Num types.
The importance of all this, ultimately, has to do with the ground types (where type class constraints are most commonly seen in functions). a represents a type, with the notation Num a => a telling us that a is a type that includes an instance of the type class Num. a could be any of the types that include the instance (ex. Int, Natural). Thus, if we give a value a general type Num a => a, we know it can implement functions for every type where there is a type class defined. For example:
ghci>> let a = 3 :: (Num a => a)
ghci>> a / 2
1.5
Whereas if we'd defined a as a specific type or in terms of a more constrained type class, we would have not been able to expect the same results:
ghci>> let a = 3 :: Integral a => a
ghci>> a / 2
-- Error: ambiguous type variable
or
ghci>> let a = 3 :: Integer
ghci>> a / 2
-- Error: No instance for (Fractional Integer) arising from a use of ‘/’
(Edit responding to followup question)
This is definitely not the most concrete explanation, so readers feel free to suggest something more rigorous.
Suppose we have a function a that is just a type class constrained version of the id function:
a :: Num a => a -> a
a = id
Let's look at type signatures for some applications of the function:
ghci>> :t (a 3)
(a 3) :: Num a => a
ghci>> :t (a 3.2)
(a 3.2) :: Fractional a => a
While our function had the general type signature, as a result of its application the the type of the application is more restricted.
Now, let's look at the function fromIntegral :: (Num b, Integral a) => a -> b. Here, the return type is the general Num b, and this will be true regardless of input. I think the best way to think of this difference is in terms of precision. fromIntegral takes a more constrained type and makes it less constrained, so we know we'll always expect the result will be constrained by the type class from the signature. However, if we give an input constraint, the actual input could be more restricted than the constraint and the resulting type would reflect that.
The reason why this works comes down to the way universal quantification works. To help explain this I am going to add in explicit forall to the type signatures (which you can do yourself if you enable -XExplicitForAll or any other forall related extension), but if you just removed them (forall a. ... becomes just ...), everything will work fine.
The thing to remember is that when a function involves a type constrained by a typeclass, then what that means is that you can input/output ANY type within that typeclass, so it's actually better to have a less constrained typeclass.
So:
fromInteger :: forall a. Num a => Integer -> a
fromInteger 5 :: forall a. Num a => a
Means that you have a value that is of EVERY Num type. So not only can you use it in a function taking it in a Fractional, you could use it in a function that only takes in MyWeirdTypeclass a => ... as long as there is one single type that implements both Num and MyWeirdTypeclass. Hence why you can get the following just fine:
fromInteger 5 / 2 :: forall a. Fractional a => a
Now of course once you decide to divide by 2, it now wants the output type to be Fractional, and thus 5 and 2 will be interpreted as some Fractional type, so we won't run into issues where we try to divide Int values, as trying to make the above have type Int will fail to type check.
This is really powerful and awesome, but very much unfamiliar, as generally other languages either don't support this, or only support it for input arguments (e.g print in most languages can take in any printable type).
Now you may be curious when the whole superclass / subclass stuff comes into play, so when you are defining a function that takes in something of type Num a => a, then because a user can pass in ANY Num type, you are correct that in this situation you cannot use functions defined on some subclass of Num, only things that work on ALL Num values, like *:
double :: forall a. Num a => a -> a
double n = n * 2 -- in here `n` really has type `exists a. Num a => a`
So the following does not type check, and it wouldn't type check in any language, because you don't know that the argument is a Fractional.
halve :: Num a => a -> a
halve n = n / 2 -- in here `n` really has type `exists a. Num a => a`
What we have up above with fromInteger 5 / 2 is more equivalent to the following, higher rank function, note that the forall within parenthesis is required, and you need to use -XRankNTypes:
halve :: forall b. Fractional b => (forall a. Num a => a) -> b
halve n = n / 2 -- in here `n` has type `forall a. Num a => a`
Since this time you are taking in EVERY Num type (just like the fromInteger 5 you were dealing with before), not just ANY Num type. Now the downside of this function (and one reason why no one wants it) is that you really do have to pass in something of EVERY Num type:
halve (2 :: Int) -- does not work
halve (3 :: Integer) -- does not work
halve (1 :: Double) -- does not work
halve (4 :: Num a => a) -- works!
halve (fromInteger 5) -- also works!
I hope that clears things up a little. All you need for the fromInteger 5 / 2 to work is that there exists ONE single type that is both a Num and a Fractional, or in other words just a Fractional, since Fractional implies Num. Type defaulting doesn't help much with clearing up this confusion, as what you may not realize is that GHC is just arbitrarily picking Double, it could have picked any Fractional.
How is a function defined for different types in Haskell?
Given
func :: Integral a => a -> a
func x = x
func' :: (RealFrac a , Integral b) => a -> b
func' x = truncate x
How could they be combined into one function with the signature
func :: (SomeClassForBoth a, Integral b) => a -> b
With a typeclass.
class TowardsZero a where towardsZero :: Integral b => a -> b
instance TowardsZero Int where towardsZero = fromIntegral
instance TowardsZero Double where towardsZero = truncate
-- and so on
Possibly a class with an associated type family constraint is closer to what you wrote (though perhaps not closer to what you had in mind):
{-# LANGUAGE TypeFamilies #-}
import GHC.Exts
class TowardsZero a where
type RetCon a b :: Constraint
towardsZero :: RetCon a b => a -> b
instance TowardsZero Int where
type RetCon Int b = Int ~ b
towardsZero = id
instance TowardsZero Double where
type RetCon Double b = Integral b
towardsZero = truncate
-- and so on
This is known as ad hoc polymorphism, where you execute different code depending on the type. The way this is done in Haskell is using typeclasses. The most direct way is to define a new class
class Truncable a where
trunc :: Integral b => a -> b
And then you can define several concrete instances.
instance Truncable Integer where trunc = fromInteger
instance Truncable Double where trunc = truncate
This is unsatisfying because it requires an instance for each concrete type, when there are really only two families of identical-looking instances. Unfortunately, this is one of the cases where it is hard to reduce boilerplate, for technical reasons (being able to define "instance families" like this interferes with the open-world assumption of typeclasses, among other difficulties with type inference). As a hint of the complexity, note that your definition assumes that there is no type that is both RealFrac and Integral, but this is not guaranteed -- which implementation should we pick in this case?
There is another issue with this typeclass solution, which is that the Integral version doesn't have the type
trunc :: Integral a => a -> a
as you specified, but rather
trunc :: (Integral a, Integral b) => a -> b
Semantically this is not a problem, as I don't believe it is possible to end up with some polymorphic code where you don't know whether the type you are working with is Integral, but you do need to know that when it is, the result type is the same as the incoming type. That is, I claim that whenever you would need the former rather than the latter signature, you already know enough to replace trunc by id in your source. (It's a gut feeling though, and I would love to be proven wrong, seems like a fun puzzle)
There may be performance implications, however, since you might unnecessarily call fromIntegral to convert a type to itself, and I think the way around this is to use {-# RULES #-} definitions, which is a dark scary bag of complexity that I've never really dug into, so I don't know how hard or easy this is.
I don't recommend this, but you can hack at it with a GADT:
data T a where
T1 :: a -> T a
T2 :: RealFrac a => a -> T b
func :: Integral a => T a -> a
func (T1 x) = x
func (T2 x) = truncate x
The T type says, "Either you already know the type of the value I'm wrapping up, or it's some unknown instance of RealFrac". The T2 constructor existentially quantifies a and packs up a RealFrac dictionary, which we use in the second clause of func to convert from (unknown) a to b. Then, in func, I'm applying an Integral constraint to the a which may or may not be inside the T.
I'm just starting with Haskell, and I thought I'd start by making a random image generator. I looked around a bit and found JuicyPixels, which offers a neat function called generateImage. The example that they give doesn't seem to work out of the box.
Their example:
imageCreator :: String -> IO ()
imageCreator path = writePng path $ generateImage pixelRenderer 250 300
where pixelRenderer x y = PixelRGB8 x y 128
when I try this, I get that generateImage expects an Int -> Int -> PixelRGB8 whereas pixelRenderer is of type Pixel8 -> Pixel8 -> PixelRGB8. PixelRGB8 is of type Pixel8 -> Pixel8 -> Pixel8 -> PixelRGB8, so it makes sense that pixelRenderer is doing some type inference to determine that x and y are of type Pixel8. If I define a type signature that asserts that they are of type Int (so the function gets accepted by generateImage, PixelRGB8 complains that it needs Pixel8s, not Ints.
Pixel8 is just a type alias for Word8. After some hair pulling, I discovered that the way to convert an Int to a Word8 is by using fromIntegral.
The type signature for fromIntegral is (Integral a, Num b) => a -> b. It seems to me that the function doesn't actually know what you want to convert it to, so it converts to the very generic Num class. So theoretically, the output of this is a variable of any type that fits the type class Num (correct me if I'm mistaken here--as I understand it, classes are kind of like "interfaces" where types are more like classes/primitives in OOP). If I assign a variable
let n = fromIntegral 5
:t n -- n :: Num b => b
So I'm wondering... what is 'b'? I can use this variable as anything, and it will implicitly cast to any numeric type, as it seems. Not only will it implicitly cast to a Word8, it will implicitly cast to a Pixel8, meaning fromPixel effectively gets turned from (as I understood it) (Integral a, Num b) => a -> b to (Integral a) => a -> Pixel8 depending on context.
Can someone please clarify exactly what's happening here? Why can I use a generic Num as any type that fits Num, both mechanically and "ethically"? I don't understand how the implicit conversion is implemented (if I were to create my own class, I feel like I would need to add explicit conversion functions). I also don't really know why this works; here I can use a pretty unsafe type and convert it implicitly to anything else. (for example, fromIntegral 50000 gets translated to 80 if I implicitly convert it to a Word8)
A common implementation of type classes such as Num is dictionary-passing. Roughly, when the compiler sees something like
f :: Num a => a -> a
f x = x + 2
it transforms it into something like
f :: (Integer -> a, a -> a -> a) -> a -> a
-- ^-- the "dictionary"
f (dictFromInteger, dictPlus) x = dictPlus x (dictFromInteger 2)
The latter basically says: "pass me an implementation for these methods of class Num for your type a, and I will use them to produce a function a -> a for you".
Values such as your n :: Num b => b are no different. They are compiled into things such as
n :: (Integer -> b) -> b
n dictFromInteger = dictFromInteger 5 -- roughly
As you can see, this turns innocent-looking integer literals into functions, which can (and does) impact performance. However, in many circumstances the compiler can realize that the full polymorphic version is not actually needed, and remove all the dictionaries.
For instance, if you write f 3 but f expects Int, the "polymorphic" 3 can be converted at compile time. So type inference can aid the optimization phase (and user-written type annotation can greatly help here). Further, some other optimizations can be triggered manually, e.g. using the GHC SPECIALIZE pragma. Finally, the dreaded monomorphism restriction tries hard to force non-functions to remain non-functions after translation, at the cost of some loss of polymorphism. However, the MR is now being regarded as harmful, since it can cause puzzling type errors in some contexts.