Haskell无法将预期类型“Integer”与实际类型“[Integer]”匹配

时间:2013-11-27 16:23:50

标签: haskell

这是我的代码,我怎么能说在2..10

之间n是int
   data Rank = Numeric Integer | Jack | Queen | King | Ace
        deriving (Eq, Show)

valueRank :: Rank ->Integer
valueRank rank

|rank ==Jack = 10
|rank ==King = 10
|rank ==Queen = 10
|rank ==Ace = 10
|rank == Numeric n = n
  where n =[x|x<-[2..10]]

1 个答案:

答案 0 :(得分:6)

我建议你使用模式匹配而不是守卫:

valueRank :: Rank -> Integer
valueRank Jack = 10
valueRank King = 10
valueRank Queen = 10
valueRank Ace = 10
valueRank (Numeric n) = n

如果您想确保无法使用超出特定范围的值创建数字,那么在进行排名时,您应使用验证此属性的smart constructor

makeRank n
  | 1 <= n <= 13 = ...
  | otherwise = error ...