QuickCheck:根据其他Arbitraries定义任意实例

Question

我正在使用QuickCheck 1,我有以下数据类型:

data A = ...
instance Arbitrary A where ...
data B = ...
instance Arbitrary B where ...
data C = C A B

现在,我想定义一个Arbitrary实例C,使C值是使用现有的发电机产生A和B.我最终这样做了:

instance Arbitrary C where
  arbitrary = elements [(C a b) |
                        a <- generate 20 (System.Random.mkStdGen 0) arbitrary,
                        b <- generate 20 (System.Random.mkStdGen 0) arbitrary]

这是明确生成固定数量的值A和B必要的,还是有更好的方法将现有的结合Arbitraries到一个新的？

Answer 1

我这样做:

instance Arbitrary C
  where arbitrary = do a <- arbitrary
                       b <- arbitrary
                       return (C a b)

虽然sclv liftM2从Control.Monad 使用的想法可能更好:

instance Arbitrary C
  where arbitrary = liftM2 C arbitrary arbitrary