Question

我有以下错误，请记住“Booleano”对应于用户定义类型。

Prop.hs:173:28: error:
    • No instance for (Foldable ((->) [Booleano]))
        arising from a use of ‘null’
    • In the first argument of ‘(==)’, namely ‘(null aux2)’
      In the expression: ((null aux2) == True)
      In the expression:
        if ((null aux2) == True) then Falso else Verdadero
Prop.hs:184:16: error:
    • No instance for (Eq Booleano) arising from a use of ‘==’
    • In the expression: x == Falso
      In an equation for ‘comparador2’: comparador2 x = x == Falso*

我正在使用此代码：

type Estados = [(String,Booleano)]

esTautologia :: Prop -> [Estados] -> Booleano
esTautologia p est =  if (null aux2 == True ) then 
              Falso
                  else
              Verdadero 
aux :: Prop -> [Estados] -> [Booleano]
aux p est= map (interp p) est

aux2:: [Booleano] -> [Booleano]
aux2= filter comparador2 aux

comparador2:: Booleano -> Bool
comparador2 x= x == Falso

类型Prop用于制作逻辑参数，因此，Prop不是关键问题。感谢您的帮助，如果您需要另外的说明，请告诉我。

Answer 1

由于filter的类型为(a -> Bool) -> [a] -> [a]，因此aux2的定义看起来很奇怪。 filter的第二个参数是一个简单的列表，而不是函数（在您的情况下为aux）。您需要更多aux参数或跳过aux才能获得正确的类型：

aux2 :: Prop -> [Estados] -> [Booleano]
aux2 a b = filter comparador2 (aux a b)


aux2 :: [Booleano] -> [Booleano]
aux2 = filter comparador2

Haskell-使用'null'时没有（可折叠（（ - ＆gt;）[Type]）的实例

1 个答案: