Question

在Haskell中如何能够解决（不使用函数而不是类型）：

重写（真假）=假重写（和（和真假）真）=假 ...

我尝试了以下

data MyLogic f a = And f a  deriving Show

rewrite(And a b)
 | a == False = False
 | b == False = False
 | otherwise = True

rewrite(And (And a b) c) = ...

但是haskell编译器抱怨在第一次重写时可能不是bool。

Answer 1

具有相同名称的多个函数定义只是ad-hoc多态的另一个名称，而这正是Haskell类型类的用途。

{-# LANGUAGE FlexibleInstances, FlexibleContexts #-}

data MyLogic f a = And f a  deriving Show

class Rewritable a where
   rewrite :: a -> Bool

instance Rewritable (MyLogic Bool Bool) where
  rewrite(And a b)
     | a == False = False
     | b == False = False
     | otherwise = True


instance Rewritable (MyLogic a b) => Rewritable  (MyLogic (MyLogic a b)  Bool) where
  rewrite(And x c) = rewrite $ And (rewrite x) c

Answer 2

以下是Hans Lub的方法的概括（完成，实际上），它应该适用于您正在考虑的那种表达方式。

{-# LANGUAGE FlexibleInstances, FlexibleContexts #-}
module MyLogic where

data MyLogic f a = And f a  deriving Show

class Rewritable a where
   rewrite :: a -> Bool

instance Rewritable Bool where
  rewrite = id

instance Rewritable (MyLogic Bool Bool) where
  rewrite(And a b) = a && b

instance Rewritable (MyLogic a b)
    => Rewritable  (MyLogic (MyLogic a b)  Bool) where
  rewrite(And x c) = rewrite $ And (rewrite x) c

instance Rewritable (MyLogic a b)
    => Rewritable  (MyLogic Bool (MyLogic a b)) where
  rewrite(And c x) = rewrite $ And c (rewrite x)

instance (Rewritable (MyLogic a b), Rewritable (MyLogic c d))
    => Rewritable (MyLogic (MyLogic a b) (MyLogic c d)) where
  rewrite(And x y) = rewrite $ And (rewrite x) (rewrite y)

Answer 3

这个怎么样：

data MyLogic = Lit Bool
             | And MyLogic MyLogic deriving Show

rewrite :: Mylogic -> Bool
rewrite (Lit b) = b
rewrite (And a b) = rewrite a && rewrite b

Answer 4

我会像这样构建它

data MyLogic a = And (MyLogic a) a  deriving Show

rewrite :: (MyLogic Bool) -> Bool
rewrite (And f False) = False
rewrite (And f True) = rewrite f

具有相同功能名称的多个函数定义

4 个答案: