或接受两个值参数haskell

Question

我是haskell的初学者，我试图为埃拉托色尼的筛定义一个简单的函数，但它说错误:

• Couldn't match expected type ‘Bool -> Bool’
with actual type ‘Bool’
• The function ‘or’ is applied to two value arguments,
but its type ‘t Bool -> Bool’ has only one
In the expression: or (mod n x) (divl l x)
In an equation for ‘divl’: divl (n : l) x = or (mod n x) (divl l x)
|
13 | divl (n:l) x = or (mod n x) (divl l x)

erat l [] = l
erat l (x:t) = if divl l x then erat l t else erat (x:t) l
divl (n:l) x = or (mod n x) (divl l x)
divl [] x = True

我试着用" ' "但是没有任何效果

Answer 1

or不是布尔或运算符;这是(||)。or接受一个布尔值列表(好吧，Foldable)，如果至少有一个值是True，则返回True。

> or []
False
> or [True, False]
True

(注意，or []被定义为False，以保持or xs || or ys == or (xs ++ ys)的标识。作为具体的例子，or [] || or [False] == or [False].)

Answer 2

尝试使用(||) :: Bool -> Bool -> Bool代替or :: Foldable t => t Bool -> Bool

https://hackage.haskell.org/package/base-4.17.0.0/docs/Prelude.html v: -124 - 124,

https://hackage.haskell.org/package/base-4.17.0.0/docs/Prelude.html v:或

。mod n x || divl l x或or [mod n x, divl l x].

Answer 3

您需要检查模是否为零，或者剩余的一个，您使用(||) :: Bool -> Bool -> Bool:

divl (n:l) x = mod n x == 0 || divl l x
divl [] x = False

基例应该是False，否则它总是返回True。但是，使用any可能更好:

divl ls x = any ((0 ==) . (`mod` x)) ls

Answer 4

最后总结了我解决的问题

erat :: [Int] -> [Int] -> [Int]
erat l [] = l
erat l (x:t) = if divl l x then erat l t else erat (x:l) t
divl (n:l) x = ((mod x n) == 0) || (divl l x)
divl [] x = False