鉴于以下数据类型:
data JoinList m a = Empty
| Single m a
| Append m (JoinList m a) (JoinList m a)
deriving (Eq, Show)使用ghci,我做到了:
*JoinList> :t Single 5 3
Single 5 3 :: (Num m, Num a) => JoinList m a为什么这里需要两种Num类型?既然这两种类型都是Num,那么为什么我们不能拥有:
Single 5 3 :: (Num m) => JoinList m m
发布于 2014-11-01 01:07:14
他们不是同一类型的。
您允许在JoinList的定义中更改类型,并且在类型签名中没有断言5和3是相同的Num。
GHCi, version 7.8.3: http://www.haskell.org/ghc/ :? for help
Loading package ghc-prim ... linking ... done.
Loading package integer-gmp ... linking ... done.
Loading package base ... linking ... done.
Prelude> :l joinlist.hs
[1 of 1] Compiling JoinList ( joinlist.hs, interpreted )
Ok, modules loaded: JoinList.
Prelude> :t Single 5 3
Single 5 3 :: (Num a, Num m) => JoinList m a
Prelude> let same = Single 5 3 :: Num a => JoinList a a
Prelude> :t same
same :: Num a => JoinList a a因为文字是多态的Num值,所以它们可以是遮罩下的两种不同的具体类型。
Prelude> let diff = Single 5 3 :: JoinList Int Float
Prelude> diff
Single 5 3.0
Prelude> :t diff
diff :: JoinList Int Floathttps://stackoverflow.com/questions/26685327
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