在Haskell中使用foldr与foldr1的用法 [英] foldr vs foldr1 usage in Haskell
问题描述
如果我写:
>让xs = [1,5,19,2,-3,5]
>最大0 xs
19
> foldr1 max xs
19
如果我写(我知道,初始值不正确这里是一个泛型最大函数...):
>让最大值'= foldr max 0
>最大'xs
19
但是,如果我写道:
>让maximum2'= foldr1 max
>最大2'xs
回应是:
< interactive>:61:11:
无法与实际类型为'Integer'的$(b)匹配预期类型
期望类型:[()]
实际类型:[Integer]
在maximum2'的第一个参数中,即`xs'
在表达式中:maximum2'xs
我是Haskell的新手。我究竟做错了什么? (无法解密错误信息...)如何使用 foldr1
和 max
?
编辑(接受答案后):
为了展示更多违约规则的例子,效果(答案也解释了这些):
示例1:
> let max'= max
> :t max
max :: Ord a => a - > a - > a
> :t max'
max'::() - > () - > ()
示例2:
>让加=(+)
> :t(+)
(+):: Num a => a - > a - > a
> :t加
plus ::整数 - >整数 - >整数
问题是关于类型和GHCi的违约。 max
的类型是多态的:
> :t max
max :: Ord a => a - > a - > a
在最大值
的情况下,编译器可以看到 a
是某种编号,而GHCi默认编号为整数:
> :t最大值'
最大值':: [Integer] - >整数
在 maximum2'
的情况下线索较少,并且默认 a
为单位类型:
> ; :t maximum2'
maximum2':: [()] - > ()
如果您提供了一个类型签名,一切都很好:
>让maximum3':: Ord a => [a] - >一个 ; maximum3'= foldr1 max
> :t maximum3'
maximum3':: Ord a => [a] - > a
我认为GHCi的违约规则可以使其他类型的情况更容易省略 - 请参阅 http://www.haskell.org/ghc/docs /7.6.2/html/users_guide/interactive-evaluation.html#id484837
If I write:
> let xs = [1,5,19,2,-3,5]
> foldr max 0 xs
19
> foldr1 max xs
19
And if I write (I know, the initial value is incorrect here for a generic maximum function...):
> let maximum' = foldr max 0
> maximum' xs
19
But if I write:
> let maximum2' = foldr1 max
> maximum2' xs
the response is:
<interactive>:61:11:
Couldn't match expected type `()' with actual type `Integer'
Expected type: [()]
Actual type: [Integer]
In the first argument of maximum2', namely `xs'
In the expression: maximum2' xs
I am new to Haskell. What am I doing wrong? (Can't decipher the error message...) How to use foldr1
with max
? Thanks.
EDIT (AFTER ACCEPTING ANSWER):
Just to show some more examples of the defaulting rules' effect (the answer explains these, too):
Example 1:
> let max' = max
> :t max
max :: Ord a => a -> a -> a
> :t max'
max' :: () -> () -> ()
Example 2:
> let plus = (+)
> :t (+)
(+) :: Num a => a -> a -> a
> :t plus
plus :: Integer -> Integer -> Integer
The problem is to do with the polymorphism of the types and GHCi's defaulting. The type of max
is polymorphic:
> :t max
max :: Ord a => a -> a -> a
In the case of maximum'
, the compiler can see that a
is some sort of number, and GHCi defaults the number to Integer:
> :t maximum'
maximum' :: [Integer] -> Integer
In the case of maximum2'
it has less clues, and defaults a
to the unit type:
> :t maximum2'
maximum2' :: [()] -> ()
If you provide a type signature, all is well:
> let maximum3' :: Ord a => [a] -> a ; maximum3' = foldr1 max
> :t maximum3'
maximum3' :: Ord a => [a] -> a
I think GHCi's defaulting rules are there to make certain other cases where the types are omitted easier -- see http://www.haskell.org/ghc/docs/7.6.2/html/users_guide/interactive-evaluation.html#id484837
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