我不确定我是否理解haskell中foldl函数的类型定义

问题描述

  code>但我甚至不确定语法。例如，为什么我不能用可折叠a ？
 
 $ b替换 ta  $ b 
还有，奖金问题，如果我必须自己定义 foldl ，我会从基本案例开始
  foldl fb [] = [] 
  
基本情况需要一个列表，比接受 Foldable 没有多大意义。什么是我可以用作基本案例的空可折叠？
       code>被称为类型类。说可折叠t => 声明 t 必须实现 Foldable 。但是， t 仍然是 t ，并且不会折叠为对可折叠界面就像在Java中一样。这就是为什么你不能只有可折叠a 。
 
 
 
绝对检查，了解 Foldable 的要求以及可以获取您的方法。
 
 
 
无论如何，如果你想使用 Foldable 的方法之一，那么可以使用已知的 Foldable 类型，比如 Set ：
 将限定数据集导入为S 
导入Data.Foldable 
 
 sumSet ::（Num n）=> S.Set n  - > n 
 sumSet ns = foldl'（\ n acc  - > n + acc）0 ns  - 如果您想要
  
或者您可以使用一个类型参数并将其限制为可折叠： 
 
 
  sumFld ::（Num n，Foldable f）=> f n  - > n  - 不同签名
 sumFld ns = foldl'（\ n acc  - > n + acc）0 ns  - 同一个实现！ 
  
以下打印6次，两次： 
 
 
  main :: IO（）
 main = print（sumSet $ S.fromList [1，2，3]）>> print（sumFld $ S.fromList [1，2，3]）
  
 
When I ask about the foldl type, what I see is this:
*Main> :t foldl
foldl :: Foldable t => (b -> a -> b) -> b -> t a -> b
In this case, what is t a? 
I guess it means that the function is using a Foldable parametrized with a but I'm not even really sure about the syntax. For instance, why can't I substitute t a with Foldable a?
And, bonus question, if I had to define foldl myself, I would start with the base case
foldl f b [] = []
But if the base case takes a list, than it would not make much sense to accept a Foldable. What is the "empty Foldable" that I could use as the base case?
 解决方案 
Foldable is something called a "typeclass".  Saying Foldable t => declares that t must implement the requirements of Foldable.  However, t remains t, and doesn't get collapsed into a reference to a Foldable interface like it might in Java.  That's why you can't just have Foldable a.
Definitely check out https://hackage.haskell.org/package/base-4.10.1.0/docs/Data-Foldable.html for an explanation of the requirements of Foldable and what methods that gets you.
Anyways, if you want to use one of Foldable's methods, then either use a known Foldable type like Set:
import qualified Data.Set as S
import Data.Foldable

sumSet :: (Num n) => S.Set n -> n
sumSet ns = foldl' (\ n acc -> n + acc ) 0 ns --make this pointfree if you want
Or you can take a type parameter and constrain it to be foldable: 
sumFld :: (Num n, Foldable f) => f n -> n --different signature
sumFld ns = foldl' (\ n acc -> n + acc ) 0 ns --same implementation!
The following prints 6, twice:
main :: IO ()
main = print (sumSet $ S.fromList [1, 2, 3]) >> print (sumFld $ S.fromList [1, 2, 3])
                        
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