Haskell 哈斯克尔'；如果我在函数中添加另一行，则SHXT失败

我试图用Haskell的hxt包解析COLLADA文件

我一直做得很好，但我遇到了一个奇怪的错误（或者更可能是我自己的错误）

我有一个箭头，看起来像这样：

processGeometry = proc x -> do
    geometry <- atTag "geometry" -< x
    meshID <- getAttrValue "id" -< geometry
    meshName <- getAttrValue "name" -< geometry
    mesh <- atTag "mesh" -< geometry
    sources <- hasName "source" <<< getChildren -< mesh
    positionSource <- hasAttrValue "id" ("-positions" `isSuffixOf`) -< sources
    positionArray  <- processFloatSource -< positionSource
    returnA -< positionArray

这是我正在尝试解析的示例COLLADA文件：

---

为什么添加一条线会完全改变箭头的结果，而它根本不应该做任何事情；DR:如果要查找两个独立的子元素，请分别调用

getChildren

您的变量

sources

不代表所有源元素的列表。相反，它是一个单一的来源。如果您检查

源的类型

，您将看到它是

XMLTree

。因此，当您对它使用两次

hasAttrValue

时，您将寻找一个与这两种情况都匹配的源元素

至于为什么返回什么并不重要：即使没有使用每一行的值，也会执行每一行。事实上，除非您正在使用输出，否则您甚至不必为其指定名称：一行只包含

hasAttrValue“id”（isSuffixOf“-normals”）[b]}
实例类别列表箭头，其中
id=ListArrow（\x->[x]）
（箭头g）。（ListArrow f）=ListArrow（\x->concatMap g（f x））
实例Arrow ListArrow where
arr f=ListArrow（\x->[f x]）
第一个（ListArrow f）=ListArrow（\（a，b）->[（a'，b）| a'a->ListArrow（树a）（树a）
hasContent=列表箭头hc，其中
hc cur@（树c_u）=如果内容==c，则[cur]else[]
getContent:：ListArrow（树a）a
getContent=ListArrow gc其中
gc（树c)）=[c]
--这与问题中的代码有相同的问题
findBothChildrenBad:：ListArrow（树字符串）（字符串，字符串）
findBothChildrenBad=proc root->do
--child是根的（单个）子级
小孩
normalSource <- hasAttrValue "id" ("-normals" `isSuffixOf`) -< sources

atTag tag = deep (isElem >>> hasName tag)

data Tree a = Tree a [Tree a]

exampleTree :: Tree String
exampleTree = Tree "root" [Tree "childA" [], Tree "childB" []]

newtype ListArrow a b = ListArrow { runListArrow :: a -> [b] }

instance Category ListArrow where
    id = ListArrow (\x -> [x])
    (ListArrow g) . (ListArrow f) = ListArrow (\x -> concatMap g (f x))

instance Arrow ListArrow where
    arr f = ListArrow (\x -> [f x])
    first (ListArrow f) = ListArrow (\(a, b) -> [ (a', b) | a' <- f a ])

getChildren :: ListArrow (Tree a) (Tree a)
getChildren = ListArrow gc where
    gc (Tree _ children) = children

hasContent :: Eq a => a -> ListArrow (Tree a) (Tree a)
hasContent content = ListArrow hc where
    hc cur@(Tree c _) = if content == c then [cur] else []

getContent :: ListArrow (Tree a) a
getContent = ListArrow gc where
    gc (Tree c _) = [c]

-- this has the same problem as the code in the question
findBothChildrenBad :: ListArrow (Tree String) (String, String)
findBothChildrenBad = proc root -> do
    -- child is a (single) child of the root
    child <- getChildren -< root

    -- childA == child, and filter to only cases where its content is "childA"
    childA <- hasContent "childA" -< child

    -- childB == child, and filter to only cases where its content is "childB"
    childB <- hasContent "childB" -< child
    -- now the content has to be both "childA" and "childB" -- so we're stuck

    childAContent <- getContent -< childA
    childBContent <- getContent -< childB
    returnA -< (childAContent, childBContent)

-- this is the fixed version
findBothChildren :: ListArrow (Tree String) (String, String)
findBothChildren = proc root -> do
    -- childA is a (single) child of the root
    childA <- getChildren -< root

    -- filter to only cases where its content is "childA"
    hasContent "childA" -< childA

    -- childB is a (potentially different) child of the root
    childB <- getChildren -< root

    -- filter to only cases where its content is "childB"
    hasContent "childB" -< childB
    -- we're not stuck here

    childAContent <- getContent -< childA
    childBContent <- getContent -< childB
    returnA -< (childAContent, childBContent)