[英]How to use foldr correctly in Haskell?
I'm trying to write a function which behave like this: 我正在尝试编写一个行为如下的函数:
correctCards :: [Card] -> [Card] -> Int
It takes two lists of type Card and check for how many cards are the same. 它需要两个类型卡列表并检查有多少卡是相同的。 Here is my code: 这是我的代码:
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. 但是类型不匹配。 Can someone tell me where I went wrong? 谁能告诉我哪里出错了? Thanks. 谢谢。
Have a look at foldr
's type: 看看foldr
的类型:
foldr :: (a -> b -> b) -> b -> [a] -> b
Now, that means that the function you supply must be of type a -> b -> b
. 现在,这意味着您提供的函数必须是a- a -> b -> b
。 Given the following code: 给出以下代码:
foldr step 0 cards
-- cards :: [Card]
-- 0 :: Integer
-- step :: ???
what should the type of step
be? step
的类型应该是什么? Given by our arguments, a
should be Card
and b
should be Integer
: 根据我们的论点, a
应该是Card
而b
应该是Integer
:
-- concrete type of `foldr` in this case
foldr :: (Card -> Integer -> Integer) -> Integer -> [Card] -> Integer
Therefore, step
should have the type (Card -> Integer -> Integer)
. 因此, step
应该具有类型(Card -> Integer -> Integer)
。 Compare this to your step function: 将此与您的 step函数进行比较:
step acc guess
| elem (head guess) answer = acc + 1
| otherwise = acc
In this case step
is Integer -> [Card] -> Integer
. 在这种情况下, step
是Integer -> [Card] -> Integer
。 And that's not the correct type. 那不是正确的类型。 Instead, you want 相反,你想要
step guess acc
| elem guess answer = acc + 1
| otherwise = acc
Note that step
only takes a single , not a whole list. 需要注意的是step
只需要一个单一的 ,而不是整个列表。
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