如何在Haskell中正确使用foldr？ [英] How to use foldr correctly in Haskell?

问题描述

  correctCards :: [Card]  - >我试图编写一个表现如下的函数： [卡片]  - > Int 
  

它需要两张卡片清单并检查有多少张卡片相同。这是我的代码：

  correctCards答案猜测= foldr步骤acc猜测
其中
 acc = 0 
步骤acc猜测
 | elem（head guess）答案= acc + 1 
 |否则= acc 
  

但是类型不匹配。有人能告诉我我错了哪里？谢谢。

解决方案

查看 foldr 的类型：

  foldr ::（a  - > b  - > b） - > b  - > [a]  - > b 
  

现在，这意味着您提供的函数必须是 a - > b - > B'/ code>。给出以下代码：



  foldr step 0 cards 
  -  cards :: [Card] 
  -  -  0 ::整数
  -  step :: ??? 
  

step 应该是什么类型？根据我们的论点， a 应该是卡片和 b 应该是整数：

   - 具体类型的`foldr` in这种情况下
 foldr ::（Card  - > Integer  - > Integer） - >整数 - > [卡片]  - >整数
  

因此， step 应该有类型（卡 - >整数 - >整数）。将这个和你的步骤函数比较：



  step acc guess 
 | elem（head guess）答案= acc + 1 
 |否则= acc 
  

在这种情况下 step 是整数 - > [卡片] - >整数。这不是正确的类型。相反，你想



  step guess acc 
 | elem guess answer = acc + 1 
 |否则= acc 
  

请注意 step 一个单个，而不是整个列表。

I'm trying to write a function which behave like this:

correctCards :: [Card] -> [Card] -> Int

It takes two lists of type Card and check for how many cards are the same. Here is my code:

correctCards answer guess = foldr step acc guess
        where 
            acc = 0
            step acc guess
                | elem (head guess) answer  = acc + 1
                | otherwise                 = acc

But the types are not match. Can someone tell me where I went wrong? Thanks.

解决方案

Have a look at foldr's type:

foldr :: (a -> b -> b) -> b -> [a] -> b

Now, that means that the function you supply must be of type a -> b -> b. Given the following code:

foldr step 0 cards
-- cards :: [Card]
-- 0     :: Integer
-- step  :: ???

what should the type of step be? Given by our arguments, a should be Card and b should be Integer:

-- concrete type of `foldr` in this case
foldr :: (Card -> Integer -> Integer) -> Integer -> [Card] -> Integer

Therefore, step should have the type (Card -> Integer -> Integer). Compare this to your step function:

step acc guess
    | elem (head guess) answer  = acc + 1
    | otherwise                 = acc

In this case step is Integer -> [Card] -> Integer. And that's not the correct type. Instead, you want

step guess acc
     | elem guess answer = acc + 1
     | otherwise         = acc

Note that step only takes a single, not a whole list.

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