有多少个参数需要Haskell的foldr函数？ [英] How many arguments takes the foldr function of Haskell?

问题描述

我是新来的Haskell，我读的书真世界哈斯克尔。在本书的第4章中，作者要求使用fold来重写groupBy函数。这本书的读者之一（Octavian Voicu）给出了以下解决方案：

  
 theCoolGroupBy ::（a  - >一 - >布尔） - >并[a]  - > [[α]] 
 theCoolGroupBy当量XS =尾$ foldr相似步骤（\_  - > [[]]）XS $（\_  - >假）
其中step x acc = \ p  - > if px then rest p else []：rest（eq x）
 where rest q = let y：ys = acc q in（x：y） ：ys 
  

我的问题很简单：我知道foldr需要3个参数：一个函数，一个初始值和一个列表。但是在代码的第二行中，foldr需要4个参数。 为什么发生这种情况？
谢谢。

解决方案

Scott的答案是正确的， foldr 是一个函数，所以这就是为什么看起来 foldr 需要4个参数。 foldr 函数确实需要3个参数（函数，基数，列表）：

  *主> ：type foldr 
 foldr ::（a  - > b  - > b） - > b  - > [a]  - > b 
  

我会在这里举一个简单的例子：

  inc :: Int  - > （Int→Int）
 inc v = \ x  - > x + v 
 
 test = inc 2 40  - 测试输出为42 
  

在上面的代码中， inc 接受一个参数， v ，并返回一个函数， code> v 。

如下所示， inc 2


  * Main> ：type inc 
 inc :: Int  - > Int  - > Int 
 * Main> ：type inc 2 
 inc 2 :: Int  - > Int 
 * Main> ：type inc 2 40 
 inc 2 40 :: Int 
  

圆括号可以用来强调返回值是一个函数，但在功能上与上面的代码相同：



  * Main> （inc 2）40 
 42 
  

PS：我是原评论的作者：）


I am new to Haskell and I am reading the book "Real World Haskell". In the Chapter 4 of the book the author asks as an exercise to rewrite the groupBy function using fold. One of the readers of the book (Octavian Voicu ) gave the following solution:

theCoolGroupBy :: (a -> a -> Bool) -> [a] -> [[a]]
theCoolGroupBy eq xs = tail $ foldr step (\_ -> [[]]) xs $ (\_ -> False)
                       where step x acc = \p -> if p x then rest p else []:rest (eq x)
                                          where rest q = let y:ys = acc q in (x:y):ys

My question is simple: I know that foldr takes 3 arguments: a function, an initial value and a list. But in the second line of the code foldr takes 4 arguments. Why this happens? Thank you.

解决方案

Scott's answer is correct, the result of the foldr is a function, so this is why it seems that foldr takes 4 arguments. The foldr functions does take 3 arguments (function, base, list):

*Main> :type foldr
foldr :: (a -> b -> b) -> b -> [a] -> b

I'll just give here an example that is less complex:

inc :: Int -> (Int -> Int)
inc v = \x -> x + v

test = inc 2 40  -- output of test is 42

In the above code, inc takes one argument, v, and returns a function that increments its argument by v.

As we can see below, the return type of inc 2 is a function, so its argument can simply be added at the end:

*Main> :type inc
inc :: Int -> Int -> Int
*Main> :type inc 2
inc 2 :: Int -> Int
*Main> :type inc 2 40                                                        
inc 2 40 :: Int

Parentheses could be used to emphasize that the return value is a function, but functionally it is identical to the above code:

*Main> (inc 2) 40
42

PS: I'm the author of the original comment :)

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