创建一个三元逻辑表,我想为我称之为<=>
的运算符创建自己的函数。
所以,例如,我想这样做,但那是不对的。这样做的正确方法是什么?
data Ternary = T | F | M
deriving (Eq, Show, Ord)
<=> :: Ternary -> Ternary -> Ternary
<=> T F = F
<=> T T = T
<=> T M = M
<=> F F = T
<=> F T = F
<=> F M = M
<=> M F = M
<=> M T = M
<=> M M = T
答案 0 :(得分:42)
只需在运营商周围添加括号:
(<=>) :: Ternary -> Ternary -> Ternary
(<=>) T F = F
(<=>) T T = T
(<=>) T M = M
(<=>) F F = T
(<=>) F T = F
(<=>) F M = M
(<=>) M F = M
(<=>) M T = M
(<=>) M M = T
这将它从中缀形式转换为前缀形式。或者,您可以在定义中使用中缀:
(<=>) :: Ternary -> Ternary -> Ternary
T <=> F = F
T <=> T = T
T <=> M = M
F <=> F = T
F <=> T = F
F <=> M = M
M <=> F = M
M <=> T = M
M <=> M = T
答案 1 :(得分:10)
带符号的函数名与不带符号的函数名不同:
-- Works:
(<^>) :: Int -> Int -> Int
a <^> b = a + b
-- Doesn't work:
{-
<^> :: Int -> Int -> Int
<^> a b = a + b
-}
-- Works:
letters :: Int -> Int -> Int
letters a b = a + b
-- Doesn't work:
{-
(letters) :: Int -> Int -> Int
a letters b = a + b
-}
我保证,Haskell非常值得学习复杂的规则。
答案 2 :(得分:1)
您可以按如下方式简化(按行)定义:
(<=>) :: Ternary -> Ternary -> Ternary
T <=> T = T
F <=> F = T
M <=> M = T
M <=> _ = M
_ <=> M = M
_ <=> _ = F
答案 3 :(得分:0)
由于您有Eq
和Ord
,因此您可以执行以下操作:
data Ternary = T | F | M
deriving (Eq, Show, Ord)
(<=>) :: Ternary -> Ternary -> Ternary
x <=> y = if x == y then T else max x y
如果您碰巧将其更改为M <=> M == M
,那么您可以执行以下操作:
data Ternary = M | T | F
deriving (Eq, Show, Ord, Enum)
(<=>) :: Ternary -> Ternary -> Ternary
x <=> y = fromEnum $ rem (toEnum x * toEnum y) 3