Учтите следующее:
import Test.QuickCheck
import Test.QuickCheck.Checkers
import Test.QuickCheck.Classes
data List a = Nil | Cons a (List a) deriving (Eq, Show)
instance Functor List where
fmap _ Nil = Nil
fmap f (Cons a l) = Cons (f a) (fmap f l)
instance Eq a => EqProp (List a) where (=-=) = eq
genList :: Arbitrary a => Gen (List a)
genList = do
n <- choose (3 :: Int, 5)
gen <- arbitrary
elems <- vectorOf n gen
return $ build elems
where build [] = Nil
build (e:es) = Cons e (build es)
instance Arbitrary a => Arbitrary (List a) where
arbitrary = frequency [ (1, return Nil)
, (3, genList)
]
main = quickBatch $ functor (Nil :: List (Int, String, Int))
Это не компилируется из-за двусмысленности в genList
:
• Could not deduce (Arbitrary (Gen a))
arising from a use of ‘arbitrary’
from the context: Arbitrary a
bound by the type signature for:
genList :: forall a. Arbitrary a => Gen (List a)
at reproducer.hs:13:1-38
• In a stmt of a 'do' block: gen <- arbitrary
In the expression:
do n <- choose (3 :: Int, 5)
gen <- arbitrary
elems <- vectorOf n gen
return $ build elems
In an equation for ‘genList’:
genList
= do n <- choose (3 :: Int, 5)
gen <- arbitrary
elems <- vectorOf n gen
....
where
build [] = Nil
build (e : es) = Cons e (build es)
|
16 | gen <- arbitrary
| ^^^^^^^^^
Я знаю, что могу написать genList
как genList = Cons <$> arbitrary <*> arbitrary
, но мне хотелось бы знать. Как убрать двусмысленность? Разве утверждение типа в main
не должно очистить его?
let gen = arbitrary
? (илиelems <- vectorOf n arbitrary
). - person Lee   schedule 04.01.2018gen
не следует разворачивать. - person Jan Synáček   schedule 04.01.2018