解析错误输入'appendString'Haskell

Question

嘿家伙，所以这是一个奇怪的小错误，我得到了，我不明白为什么它给了我。

它说，在输入解析错误“appendString”但我认为没有错吧......

我把它从若，然后else语句如下：

createShow currentIndex (Grid {delta = d, middle = (x,y), points = g}) dir counter = 
if currentIndex ==0 || dir == 2 
    then (appendString d (x,y) g currentIndex) ++ (createShow currentIndex+1 (Grid {delta = d, middle = (x,y), points = g}) 2 (counter+1)) 
else if counter == (2*d+1) 
    then (appendString d (x,y) g currentIndex) ++ (appendX x) 
else if dir == 1 
    then (appendString d (x,y) g currentIndex) ++ (createShow currentIndex-1 (Grid {delta = d, middle = (x,y), points = g}) 1 (counter+1)) 

其中createShow回报一个字符串，所以没有appendString

appendString给构造函数中的错误：

appendString d (x,y) g currentIndex = 
(if currentIndex == y 
    then "y " 
else 
    " ") ++ (show currentIndex) ++(rowFunction g x d 0 (x+d) 1)++ "\n" 

你看到我可能出错了吗？

编辑：增加了整个区域

Answer 1

哈斯克尔if的不是像其他if“中说Java或Python秒。最大的区别是他们是表达式不像java或python，它们是语句。

他们是从C

更接近condition ? res1 : res2写嵌套if的是这样的正确方法：

if condition 
    then foo 
    else if condition2 
     then bar 
     else ... 

你会发现，这是可怕的丑陋。

这就是为什么Haskell有警卫：

foo args | condition = foo 
     | condition2= bar 
     | otherwise = meh 

这里我们声明函数foo，如果condition为真，那么我们执行foo否则我们继续condition2和otherwise始终为true。对你而言

createShow currentIndex (Grid {delta = d, middle = (x,y), points = g}) dir counter 
    | currentIndex == 0 || dir == 2 = appendString d .... 
    | counter == (2 * d + 1)  = appendString d .... 
    | dir == 1      = appendString d .... 

看起来更具可读性。

Answer 2

这里有一个重构::

createShow currentIndex grid@(Grid {delta = d, middle = (x,y), points = g}) dir counter = 
    prefix ++ show currentIndex ++ row ++ "\n" ++ nextLine 
    where 
    prefix = if currentIndex == y then "y " else " " 
    row = rowFunction g x d 0 (x+d) 1 

    nextLine | currentIndex == 0 || dir == 2 = createShow (currentIndex+1) grid 2 (counter+1) 
    nextLine | counter == (2*d+1)   = appendX x 
    nextLine | dir == 1      = createShow (currentIndex-1) grid 1 (counter+1) 

    appendX x = ... 

事情需要注意：

• 使用where条款往往可以让你避免重复参数
• 至appendString常见的呼叫已被提取出来，并移动到顶端，然后内联，因为它只被调用一次。
• 使用警卫nextLine更清楚地处理级联if。
• 看守和形式nextLine说清楚它不是总计函数。当它结束时会发生什么？
• 使用grid@来命名模式。这样，在进行递归调用时，您不需要“重构”Grid值。

可以走得更远。我注意到Grid {...}和dir从来没有在整个函数变化表明融通那些出：

createShow currentIndex (Grid {delta = d, middle = (x,y), points = g}) dir counter = 
    line currentIndex counter 
    where 
    line currentIndex counter = 
     prefix ++ show currentIndex ++ row ++ "\n" ++ nextLine currentIndex counter 

    prefix = if currentIndex == y then "y " else " " 
    row = rowFunction g x d 0 (x+d) 1 

    nextLine currentIndex counter 
     | currentIndex == 0 || dir == 2 = line (currentIndex+1) (counter+1) 
     | counter == (2*d+1)   = appendX x 
     | dir == 1      = line (currentIndex-1) (counter+1) 

    appendX x = ... 

这里，line饰演“携带”周围的函数递归不同的唯一值的一部分。这将是一个常见的成语来放置这些参数在什么createShow需要结束，因此即使将它们剔除：

createShow :: Grid -> Int -> Int -> Int -> String 
createShow (Grid {delta = d, middle = (x,y), points = g}) dir = line 
    where 
    line currentIndex counter = 
     prefix ++ show currentIndex ++ row ++ "\n" ++ nextLine currentIndex counter 

    prefix = if currentIndex == y then "y " else " " 
    row = rowFunction g x d 0 (x+d) 1 

    nextLine currentIndex counter 
     | currentIndex == 0 || dir == 2 = line (currentIndex+1) (counter+1) 
     | counter == (2*d+1)   = appendX x 
     | dir == 1      = line (currentIndex-1) (counter+1) 

    appendX x = ...