我正在尝试使用以下代码实现luhn算法:
luhn :: Int -> Bool
luhn x = (tail $ show (foldl (\acc x -> acc + (read x :: Int)) 0 (foldr doEncrypt [] $ zip [0..] (show x)))) == "0"
where
doEncrypt (i,y) acc = if not(even i)
then head(((uncurry (+) . (`divMod` 10) . (*2)) y)) : acc
else (head y) : acc
我现在遇到了以下错误:
• Non type-variable argument in the constraint: Integral [a2]
(Use FlexibleContexts to permit this)
• When checking the inferred type
doEncrypt :: forall a1 a2.
(Integral a1, Integral [a2]) =>
(a1, [a2]) -> [a2] -> [a2]
In an equation for ‘luhn’:
luhn x
= (tail
$ show
(foldl
(\ acc x -> acc + (read x :: Int))
0
(foldr doEncrypt [] $ zip [0 .. ] (show x))))
== "0"
where
doEncrypt (i, y) acc
= if not (even i) then
head (((uncurry (+) . (`divMod` 10) . (* 2)) y)) : acc
else
(head y) : acc
我看到错误表明元组的第二部分(a2
)是“非类型变量参数”。然而,Haskell似乎将这个a2
论证确定为Integral
,而实际上它是Char
。我如何告诉Haskell这是一个Char
,Haskell不应该担心这个变量的类型?或者是否有其他我不理解导致此错误的原因?
编辑:当我删除(head y)
并用y
替换它时,我得到以下错误:
• Couldn't match type ‘Char’ with ‘[Char]’
Expected type: [String]
Actual type: [Char]
• In the third argument of ‘foldl’, namely
‘(foldr doEncrypt [] $ zip [0 .. ] (show x))’
In the first argument of ‘show’, namely
‘(foldl
(\ acc x -> acc + (read x :: Int))
0
(foldr doEncrypt [] $ zip [0 .. ] (show x)))’
In the second argument of ‘($)’, namely
‘show
(foldl
(\ acc x -> acc + (read x :: Int))
0
(foldr doEncrypt [] $ zip [0 .. ] (show x)))’
我的解决方案有多少错误,但最后以下代码有效!
luhn :: Int -> Bool
luhn x = (tail $ show (foldl (\acc x -> acc + (digitToInt x)) 0 (foldr doEncrypt [] $ zip [0..] (show x)))) == "0"
where
doEncrypt (i,y) acc = if not(even i)
then (head $ show(((uncurry (+) . (`divMod` 10) . (*2)) (digitToInt y)))) : acc
else y : acc
非常感谢@WillemVanOnsem的指示,没有我可能不会解决这个问题!