Haskell类型类错误
我搞不懂,代码:Haskell类型类错误,haskell,Haskell,我搞不懂,代码: smallSum :: (Ord a, Integral a) => a -> a smallSum n | n < 0 = 0 | (n < 20) = n + smallSum (n - 1) | otherwise = error "Number needs to be in 1..10" fastSumOfSeriesLength :: (Ord a, Integral a) => a -> a fastS
smallSum :: (Ord a, Integral a) => a -> a
smallSum n
| n < 0 = 0
| (n < 20) = n + smallSum (n - 1)
| otherwise = error "Number needs to be in 1..10"
fastSumOfSeriesLength :: (Ord a, Integral a) => a -> a
fastSumOfSeriesLength x
| x < 10 = smallSum x
| x >= 10 = sum (take (rest - 1) [dif !! (firstDigit - 1), dif !! (firstDigit - 1) + 100..]) + smallList !! (firstDigit - 1)
where
smallList = [smallSum x | x <- [1..10]]
largeList = [smallSum x | x <- [11..20]]
dif = [l - s | l <- largeList, s <- smallList]
firstDigit = x `mod` 10
rest = x `div` 10
我期待有人指出什么是错误的,它会看起来轻工作,我需要谷歌了解更多关于这个错误 查看
(!!)
和获取的类型:
*Main> :t (!!)
(!!) :: [a] -> Int -> a
*Main> :t take
take :: Int -> [a] -> [a]
由于在表达式上使用这些函数的类型与x
相同,这意味着x
必须是Int
——但您声明此函数应适用于任何类型的(整数)数。(如果你慢慢读错误,希望你能看到它这么说。)最简单的修复方法是导入数据。列出并使用genericIndex
和genericTake
。快速提示:你可以删除函数的类型声明,然后将其加载到GHCi中。使用:t expr
命令,您可以询问函数的类型。我更愿意说,最简单的修复方法是删除类型声明或使其为Int->Int,但这取决于他是否需要多态性。@MdxBhmt可能最简单的方法是删除整个文件。但我对“简单”的标准不如我对“简单”的标准有效
*Main> :t (!!)
(!!) :: [a] -> Int -> a
*Main> :t take
take :: Int -> [a] -> [a]