While using applicative functors in Haskell I've often run into situations where I end up with repetitive code like this:
instance Arbitrary MyType where
arbitrary = MyType <$> arbitrary <*> arbitrary <*> arbitrary <*> arbitrary
In this example I'd like to say:
instance Arbitrary MyType where
arbitrary = applyMany MyType 4 arbitrary
but I can't figure out how to make applyMany
(or something similar to it). I can't even figure out what the type would be but it would take a data constructor, an Int (n) , and a function to apply n times. This happens when creating instances for QuickCheck, SmallCheck, Data.Binary, Xml serialization, and other recursive situations.
So how could I define applyMany
?
Check out derive . Any other good generics library should be able to do this as well; derive is just the one I am familiar with. For example:
{-# LANGUAGE TemplateHaskell #-}
import Data.DeriveTH
import Test.QuickCheck
$( derive makeArbitrary ''MyType )
To address the question you actually asked, FUZxxl is right, this is not possible in plain vanilla Haskell. As you point out, it is not clear what its type should even be. It is possible with Template Haskell metaprogramming (not too pleasant). If you go that route, you should probably just use a generics library which has already done the hard research for you. I believe it is also possible using type-level naturals and typeclasses, but unfortunately such type-level solutions are usually difficult to abstract over. Conor McBride is working on that problem .
I think you can do it with OverlappingInstances hack:
{-# LANGUAGE FlexibleInstances, MultiParamTypeClasses, TypeFamilies, OverlappingInstances #-}
import Test.QuickCheck
import Control.Applicative
class Arbitrable a b where
convert :: Gen a -> Gen b
instance (Arbitrary a, Arbitrable b c) => Arbitrable (a->b) c where
convert a = convert (a <*> arbitrary)
instance (a ~ b) => Arbitrable a b where
convert = id
-- Should work for any type with Arbitrary parameters
data MyType a b c d = MyType a b c d deriving (Show, Eq)
instance Arbitrary (MyType Char Int Double Bool) where
arbitrary = convert (pure MyType)
check = quickCheck ((\s -> s == s) :: (MyType Char Int Double Bool -> Bool))
Not satisfied with my other answer, I have come up with an awesomer one.
-- arb.hs
import Test.QuickCheck
import Control.Monad (liftM)
data SimpleType = SimpleType Int Char Bool String deriving(Show, Eq)
uncurry4 f (a,b,c,d) = f a b c d
instance Arbitrary SimpleType where
arbitrary = uncurry4 SimpleType `liftM` arbitrary
-- ^ this line is teh pwnzors.
-- Note how easily it can be adapted to other "simple" data types
ghci> :l arb.hs
[1 of 1] Compiling Main ( arb.hs, interpreted )
Ok, modules loaded: Main.
ghci> sample (arbitrary :: Gen SimpleType)
>>>a bunch of "Loading package" statements<<<
SimpleType 1 'B' False ""
SimpleType 0 '\n' True ""
SimpleType 0 '\186' False "\208! \227"
...
Lengthy explanation of how I figured this out
So here's how I got it. I was wondering, "well how is there already an Arbitrary
instance for (Int, Int, Int, Int)
? I'm sure no one wrote it, so it must be derived somehow. Sure enough, I found the following in the docs for instances of Arbitrary :
(Arbitrary a, Arbitrary b, Arbitrary c, Arbitrary d) => Arbitrary (a, b, c, d)
Well, if they already have that defined, then why not abuse it? Simple types that are merely composed of smaller Arbitrary data types are not much different than just a tuple.
So now I need to somehow transform the "arbitrary" method for the 4-tuple so that it works for my type. Uncurrying is probably involved.
Stop. Hoogle time!
(We can easily define our own uncurry4
, so assume we already have this to operate with.)
I have a generator, arbitrary :: Gen (q,r,s,t)
(where q,r,s,t are all instances of Arbitrary). But let's just say it's arbitrary :: Gen a
. In other words, a
represents (q,r,s,t)
. I have a function, uncurry4
, which has type (q -> r -> s -> t -> b) -> (q,r,s,t) -> b
. We are obviously going to apply uncurry4 to our SimpleType
constructor. So uncurry4 SimpleType
has type (q,r,s,t) -> SimpleType
. Let's keep the return value generic, though, because Hoogle doesn't know about our SimpleType. So remembering our definition of a
, we have essentially uncurry4 SimpleType :: a -> b
.
So I've got a Gen a
and a function a -> b
. And I want a Gen b
result. (Remember, for our situation, a
is (q,r,s,t)
and b
is SimpleType
). So I am looking for a function with this type signature: Gen a -> (a -> b) -> Gen b
. Hoogling that , and knowing that Gen
is an instance of Monad
, I immediately recognize liftM
as the monadical-magical solution to my problems.
Hoogle saves the day again. I knew there was probably some "lifting" combinator to get the desired result, but I honestly didn't think to use liftM (durrr!) until I hoogled the type signature.
Here is what I'v got at least:
{-# LANGUAGE TypeFamilies, MultiParamTypeClasses, FlexibleInstances #-}
{-# LANGUAGE FlexibleContexts #-}
module ApplyMany where
import Control.Applicative
import TypeLevel.NaturalNumber -- from type-level-natural-number package
class GetVal a where
getVal :: a
class Applicative f => ApplyMany n f g where
type Res n g
app :: n -> f g -> f (Res n g)
instance Applicative f => ApplyMany Zero f g where
type Res Zero g = g
app _ fg = fg
instance
(Applicative f, GetVal (f a), ApplyMany n f g)
=> ApplyMany (SuccessorTo n) f (a -> g)
where
type Res (SuccessorTo n) (a -> g) = Res n g
app n fg = app (predecessorOf n) (fg<*>getVal)
Usage example:
import Test.QuickCheck
data MyType = MyType Char Int Bool deriving Show
instance Arbitrary a => GetVal (Gen a) where getVal = arbitrary
test3 = app n3 (pure MyType) :: Gen MyType
test2 = app n2 (pure MyType) :: Gen (Bool -> MyType)
test1 = app n1 (pure MyType) :: Gen (Int -> Bool -> MyType)
test0 = app n0 (pure MyType) :: Gen (Char -> Int -> Bool -> MyType)
Btw, I think this solution is not very useful in real world. Especially without local type-classes.
Check out liftA2 and liftA3 . Also, you can easily write your own applyTwice or applyThrice methods like so:
applyTwice :: (a -> a -> b) -> a -> b
applyTwice f x = f x x
applyThrice :: (a -> a -> a -> b) -> a -> b
applyThrice f x = f x x x
There's no easy way I can see to get the generic applyMany you're asking for, but writing trivial helpers such as these is neither difficult nor uncommon.
[edit] So it turns out, you'd think something like this would work
liftA4 f a b c d = f <$> a <*> b <*> c <*> d
quadraApply f x = f x x x x
data MyType = MyType Int String Double Char
instance Arbitrary MyType where
arbitrary = (liftA4 MyType) `quadraApply` arbitrary
But it doesn't. (liftA4 MyType)
has a type signature of (Applicative f) => f Int -> f String -> f Double -> f Char -> f MyType
. This is incompatible with the first parameter of quadraApply, which has a type signature of (a -> a -> a -> a -> b) -> a -> b
. It would only work for data structures that hold multiple values of the same Arbitrary type.
data FourOf a = FourOf a a a a
instance (Arbitrary a) => Arbitrary (FourOf a) where
arbitrary = (liftA4 FourOf) `quadraApply` arbitrary
ghci> sample (arbitrary :: Gen (FourOf Int))
Of course you could just do this if you had that situation
ghci> :l +Control.Monad
ghci> let uncurry4 f (a, b, c, d) = f a b c d
ghci> samples <- sample (arbitrary :: Gen (Int, Int, Int, Int))
ghci> forM_ samples (print . uncurry4 FourOf)
There might be some language pragma that can shoehorn the "arbitrary" function into the more diverse data types. But that's currently beyond my level of Haskell-fu.
This is not possible with Haskell. The problem is, that your function will have a type, that depends on the numeric argument. With a type system that allows dependent types, that should be possible, but I guess not in Haskell.
What you can try is using polymorphism and tyeclasses to archieve this, but it could become hacky and you need a big bunch of extensions to satisfy the compiler.
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