Question

我正在尝试为自定义数据类型生成任意大小的元素：

newtype ZippList a = ZPL ([a], [a])
    deriving (Show)

这就是我得到的：

instance Arbitrary a => Arbitrary (ZippList a) where
arbitrary = sized zipplist where
    zipplist n = do
        firstLength <- choose(0,n)
        secondLength <- n - firstLength
        firstList <- [arbitrary :: Gen a | [1..firstLength]]
        secondList <- [arbitrary :: Gen a | [1..secondLength]]
        return $ ZPL (firstList, secondList)

然而它没有编译。编译无法生成两个a列表。如何生成任意a？

该死的，我有点忘了用_ <-实际生成这些值。抱歉这个琐碎的问题，我编码很晚。

Answer 1

这对我有用

instance Arbitrary a => Arbitrary (ZippList a) where
  arbitrary = sized zipplist where
    zipplist n = do
        firstLength <- choose (0, n)
        let secondLength = n - firstLength
        firstList <- sequence [arbitrary | _ <- [1..firstLength]]
        secondList <- sequence [arbitrary | _ <- [1..secondLength]]
        return $ ZPL (firstList, secondList)

请注意，secondLength的定义不使用monad，因此您应该使用let

QuickCheck Generator - 自定义类型的任意元素

1 个答案: