在下面的 Haskell 代码中,如何写得更简洁?是否有必要列出所有四个条件,或者可以用更紧凑的模式来概括这些条件?例如,有没有一种方法可以利用 Haskell 已经知道如何添加 float 和整数,而无需手动指定 fromIntegral?
data Signal = SignalInt Int | SignalFloat Float | Empty deriving (Show)
sigAdd :: Signal -> Signal -> Signal
sigAdd (SignalInt a) (SignalInt b) = SignalInt (a + b)
sigAdd (SignalInt a) (SignalFloat b) = SignalFloat ((fromIntegral a) + b)
sigAdd (SignalFloat a) (SignalInt b) = SignalFloat (a + (fromIntegral b))
sigAdd (SignalFloat a) (SignalFloat b) = SignalFloat (a + b)
main :: IO ()
main = do
putStrLn (show (sigAdd (SignalFloat 2) (SignalInt 5)))
最佳答案
Haskell 不知道如何添加Float
和Int
;它对于类型非常具体和明确:
Prelude> (5 :: Int) + 3.5
<interactive>:1:13:
No instance for (Fractional Int)
arising from the literal `3.5' at <interactive>:1:13-15
Possible fix: add an instance declaration for (Fractional Int)
In the second argument of `(+)', namely `3.5'
In the expression: (5 :: Int) + 3.5
In the definition of `it': it = (5 :: Int) + 3.5
定义一个函数toFloatSig
:
toFloatSig (SignalInt a) = fromIntegral a
toFloatSig (SignalFloat a) = a
然后你可以写:
sigAdd (SignalInt a) (SignalInt b) = SignalInt (a + b)
sigAdd sa sb = SignalFloat (toFloatSig sa + toFloatSig sb)
将 Signal
设为 Num
class 的实例也可能是合适的。 ,以便您可以直接使用 +
运算符添加它们。另外,您可以使类型更通用:
data (Num a) => Signal a = Signal a | Empty deriving (Show)
关于haskell - Haskell 中的模式匹配数据类型。捷径?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/1715871/