Haskell 奇怪的（对我来说）行为

Question

我正在解决 Haskell (https://wiki.haskell.org/99_questions/1_to_10) 的 99 个问题，我对问题 # 8 有疑问。

8 Problem 8
(**) Eliminate consecutive duplicates of list elements.

If a list contains repeated elements they should be replaced with a single copy of the element. The order of the elements should not be changed.

我已经用 foldr 函数成功解决了这个问题。

compress :: Eq e => [e] -> [e]
compress =  let f v [] = [v]
                f v acc 
                        | head acc == v = acc
                        | otherwise = v:acc
            in foldr f []

但是当我尝试像这样用递归解决同样的问题时：

compress' :: Eq e => [e] -> [e]
compress' = let f acc [] = acc
                f [] (x:xs) = f [x] xs
                f acc (x:xs) | x == last acc = acc ++ f acc xs
                             | otherwise = f (acc ++ [x]) xs 
            in f []

我看到了非常奇怪的行为。我看到这个函数的结果：

compress' "aaaabccaadeeee"
"aaaabcabcaabcadeabcadeabcadeabcade"

但是如果我在第

行添加断点

compress' = let f acc [] = acc

它给了我正确的结果：

ghci> compress' "aaaabccaadeeee"
"aaaabcabcaabcadeabcadeabcadeabcade"
ghci> :break 304
Breakpoint 7 activated at haskell-tut.hs:304:28-30
ghci> compress' "aaaabccaadeeee"
"aaaabcabcaabcadeabcadeabcadeStopped in Main.compress'.f, haskell-tut.hs:304:28-30
_result :: [Char] = _
acc :: [Char] = "abcade"
[haskell-tut.hs:304:28-30] ghci> :con
abcade"
ghci>

我觉得这与 Haskell 懒惰有关....这是我最好的假设。任何人都可以解释为什么我在执行期间会得到这个奇怪的结果，而在使用断点执行期间会得到正确的结果吗？

Answer 1

问题来自以下表达式：

x == last acc = acc ++ f acc xs

不需要在结果的开头附加acc字符串，所以更正应该是：

x == last acc = f acc xs

注意acc包含了你想要的正确结果，即没有连续重复的字符串，因此你可以在输入列表为[=15的断点处看到正确的结果acc :: [Char] = "abcade" =].但是当它returns时，它把前面的结果合并为acc ++ "abcade"，从中得到"aaaabcabcaabcadeabcadeabcadeabcade"

末尾的"abcade"

Haskell 奇怪的（对我来说）行为

Haskell strange (to me) behavior

haskell

computation