理解Haskell类型的问题

问题描述

所以我想让我看一个单词列表并返回一个列表，就像它一样，但
，每当它们显示为连续单词时，会进行以下替换。

其中一个例子是 you ，并将它变成 u

我给出以下内容：

  hep :: [ Word]  - > [Word] 
类型字=字符串
  

现在给我的问题是， m试图使用case表达式，这样我就不必重复代码，但我得到以下错误：

 无法匹配预期类型'Char'的实际类型为'[Char]'
在模式中：You
在另一种情况下：You - > u：hep xs 
在表达式中：{你 - > u：hep xs} 
  

来自以下代码

  hep [] = [] 
 hep [a：xs] = 
的情况a您 - > u：hep xs 
  

任何人都可以告诉我问题是什么？

编辑：

我已经添加了以下代码：

  hep [] = [[]] 
 hep（a：xs）= 
you的情况a  - > u：hep xs 
是 - > r：hep xs 
your - > ur：hep xs 
男朋友 - > bf：hep xs 
女朋友 - > gf：hep xs 
很棒 - > gr8：hep xs 
 a  - > a：hep xs 
  

现在我将如何添加一个案例，以便如果列表包含2或3个特定的词，我可以把它变成一个首字母缩写词吗？

Ex

  [By，The，你可以在这里找到你想要的东西，然后你就可以在这里找到你想要的东西了。[btw]] 
  

解决方案

重试与字符串列表匹配，但是 hep 的类型是 [Word] - > [Word] ，这与此相矛盾。这个错误信息是指这个事实。

但是我猜你实际上想要的是什么？

<$ （a：xs）= a
的情况aYou - >> u：hep xs
a - > a：hep xs

Hey guys so I'm suppose to takes a list of words and returns a list just like it butwith the following substitutions made each time they appear as consecutive words.

One example is you and turn it into u

I'm given the following:

hep :: [Word] -> [Word]
type Word = String

now what is giving me problem is that I'm trying to use case expressions so that I won't have to repeat code but I get the following error

Couldn't match expected type `Char' with actual type `[Char]'
In the pattern: "You"
In a case alternative: "You" -> "u" : hep xs
In the expression: case a of { "You" -> "u" : hep xs }

from the following code

hep [] = []
hep [a:xs] = case a of
    "You" -> "u":hep xs

Anyone tell me what the problem is?

Edit:

I have added the following code

hep [] = [[]]
hep (a:xs) = case a of
    "you" -> "u":hep xs
    "are" -> "r":hep xs
    "your" -> "ur":hep xs
    "boyfriend" -> "bf":hep xs
    "girlfriend" -> "gf":hep xs
    "great" -> "gr8":hep xs
    a -> a:hep xs

Now how would I be able to add a case so that if the list contains 2 or 3 certain words in an order, I could turn that into an acronym?

Ex

["By","The","way"] = ["btw"]

解决方案

You're trying to match against a list of a list of Strings, but the type of hep is [Word] -> [Word], which contradicts that. The error message is referring to this fact.

But I'm guessing what you actually want is this?

hep [] = []
hep (a:xs) = case a of
    "You" -> "u":hep xs
    a -> a:hep xs

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