Haskell 如何修复类似列表类型的任意实例的歧义性

考虑以下几点：

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))

---

我知道我可以把

genList

写成

genList=Cons arbitral

，但我想知道。如何消除歧义？

main

中的类型断言不应该清除它吗？

vectorOf

需要一个

gena

。因此，GHC试图为

Gen a

找到一个不存在的

任意

实例

使用任意n的

向量

或只使用

向量n

。此外，建议使用QuickCheck选择尺寸：

genList :: Arbitrary a => Gen (List a)
genList = sized $ \n ->
    elems <- vector n
    return $ build elems
  where build [] = Nil
        build (e:es) = Cons e (build es)

---

请注意，

genList=Cons arbitral

不会生成空列表，而

foldr Cons Nil listOf arbitral

会生成空列表。

您不是只想

let gen=arbitral

？（或者，

elems@Lee，我现在觉得自己很愚蠢。事实上，
gen
不应该被“拆开”。
genList :: Arbitrary a => Gen (List a)
genList = sized $ \n ->
    elems <- vector n
    return $ build elems
  where build [] = Nil
        build (e:es) = Cons e (build es)

genList :: Arbitrary a => Gen (List a)
genList = build <$> listOf arbitary
  where build = foldr Cons Nil