函数buildTree'Haskell中的非穷举模式

Question

我使用返回元组的函数。但是当我试图运行这个函数时，它给了我一个例外：函数中的非穷尽模式。

buildTree' :: String -> Tree  -> (String,Tree)
buildTree' (x:xs)  currenttree 
|null (x:xs) = ("empty", currenttree)
|isDigit x && ( take 1  xs == ['+'] ||  take 1 xs == ['-'])= buildTree' xs (Node [x] Empty Empty)
|isDigit x = buildTree' newstring1 (snd (buildrecursion (getminiexpr(x:xs)) Empty))
|elem x "+-" = buildTree' newstring (buildTree2 currenttree newtree [x])
    where newtree = (snd (buildrecursion (getminiexpr xs) Empty))
          newstring = drop (length(getminiexpr xs)) xs
          newstring1 = drop (length(getminiexpr (x:xs))) (x:xs)


getminiexpr :: String -> String 
getminiexpr input = takeWhile ( \y -> y /= '+' && y /= '-') input

Answer 1

（x:xs）不是任意列表/字符串，它是一个非空列表/字符串，其头是x，尾是xs。因此

buildTree' (x:xs)  currenttree 
   ...

只处理非空列表。使用-wall编译会警告缺少空列表大小写。所以，你需要：

buildTree' [] currenttree = ...
buildTree' (x:xs)  currenttree 
   ...

调整您的代码，我们可以删除null保护：

buildTree' [] currenttree = ("empty", currenttree)
buildTree' (x:xs)  currenttree 
   | isDigit x && ( take 1  xs == ['+'] ||  take 1 xs == ['-'])= buildTree' xs (Node [x] Empty Empty)
   ...

类似地，take检查需要的比列表的头x更多。你可以改为写：

buildTree' [] currenttree = ("empty", currenttree)
buildTree' (x1:x2:xs)  currenttree 
   | isDigit x1 && ( x2 == '+' || x2 == '-') = buildTree' (x2:xs) (Node [x1] Empty Empty)
   ...

或者甚至

buildTree' [] currenttree = ("empty", currenttree)
buildTree' (x1:x2:xs)  currenttree 
   | isDigit x1 && (x2 `elem` "+-") = buildTree' (x2:xs) (Node [x1] Empty Empty)
   ...

我不知道你的逻辑是否正确。