Question

我正在通过一些练习来学习一些Haskell（更好的）。

以下是我对BubbleSort的尝试：

bubbleSort :: [Integer] -> [Integer]
bubbleSort xs = let 
                  (state, bubbled) = bubble True xs
                in  
                  if not state
                  then bubbleSort bubbled
                  else bubbled
  where
    bubble :: Bool -> [Integer] -> (Bool, [Integer])
    bubble changed [] = (changed, []) 
    bubble changed [x] = (changed, [a])
    bubble changed (a:b:as) | compare a b == GT = (fst bubble False (a:as), b:(snd bubble False (a:as)))
                            | otherwise = (fst bubble (changed && True) (b:as), a:(snd bubble (changed && True) (b:as)))

这在最后一行（在else子句中）出错：Not in scope: 'a'

老实说，我不确定我是否只是非常疲惫，或者我错过了一些非常基本的东西，但据我所知，a应该在范围内，因为它是作为（a）的一部分传递的：b：as）模式？这不正确吗？

Answer 1

您的otherwise案例没有问题，但不是这种情况：bubble changed [x] = (changed, [a]) ... a不在范围内。

import           Data.List       (sort)
import           Test.QuickCheck

main :: IO ()
main = verboseCheck isValidSort >> verboseCheck idempotent

isValidSort, idempotent :: [Integer] -> Bool
isValidSort xs = sort xs == bubbleSort xs
idempotent  xs = bubbleSort (bubbleSort xs) == bubbleSort xs

bubbleSort :: [Integer] -> [Integer]
bubbleSort xs = let (state, bubbled) = bubble True xs
                in if not state
                   then bubbleSort bubbled
                   else bubbled
  where
    bubble :: Bool -> [Integer] -> (Bool, [Integer])
    bubble changed [] = (changed, [])
    bubble changed [x] = (changed, [x])
    bubble changed (a:b:as) | compare a b == GT = (fst $ bubble False (a:as), b:(snd $ bubble False (a:as)))
                            | otherwise = (fst $ bubble (changed && True) (b:as), a:(snd $ bubble (changed && True) (b:as)))

功能参数不在函数体的范围内？

1 个答案: