Question

在下面的Haskell代码中，如何更简洁地编写？是否有必要列出所有四个条件，还是可以通过更紧凑的模式进行总结？例如，有没有办法可以利用Haskell已经知道如何添加float和int，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)))

Answer 1

Haskell 不知道如何添加Float和Int;它是非常具体和明确的类型：

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作为Num class的实例也可能是合适的，这样您就可以直接使用+运算符添加它们。此外，您可以使类型更通用：

data (Num a) => Signal a = Signal a | Empty deriving (Show)

Haskell中的模式匹配数据类型。捷径？

1 个答案: