如何使用Haskell定义与Guards的独占分离

正如@sibi所提到的,你正在犯语法错误.但是你正在做的是通过模式匹配来实现的.

exclusiveDisjunction :: Bool -> Bool -> Bool
exclusiveDisjunction True True = False
exclusiveDisjunction True False = True 
exclusiveDisjunction False True = True 
exclusiveDisjunction False False = False

这可以简化为

exclusiveDisjunction :: Bool -> Bool -> Bool
exclusiveDisjunction True True = False
exclusiveDisjunction True False = True 
exclusiveDisjunction _ a = a

已经有了很好的答案,但我希望从我的角度展示如何以更清晰的方式实现这一点.

exclusiveDisjunction :: Bool -> Bool -> Bool
exclusiveDisjunction True = not
exclusiveDisjunction False = id

最短的.

exclusiveDisjunction :: Bool -> Bool -> Bool
exclusiveDisjunction = (/=)