将Scheme闭包定义函数转换为Haskell

Question

以下程序,用Scheme编写,

(define (eat xs)
  (lambda (x)
    (if (eq? x 'vomit)
      (reverse xs)
      (eat (cons x xs)))))

可以通过以下方式调用:

> eat 
#{procedure 9165 eat}
> (eat '())
#{procedure 9166 (unnamed in eat)}
> ((eat '()) 1)
#{procedure 9166 (unnamed in eat)}
> (((((eat '()) 1) 2) 3) 4)
#{procedure 9166 (unnamed in eat)}
> ((((((eat '()) 1) 2) 3) 4) 'vomit)
(1 2 3 4) 

因为没有突变,例如.set!,并且使用递归传递状态我认为在Haskell中编写这将是微不足道的:

eat xs x = if x == "vomit" then reverse xs else eat (x:xs)

-- Couldn't match expected type ‘[[Char]]’
--        with actual type ‘[Char] -> [[Char]]’
-- Probable cause: ‘eat’ is applied to too few arguments
-- In the expression: eat (x : xs)
-- In the expression:
--   if x == "vomit" then reverse xs else eat (x : xs)

我错过了一些明显的东西,还是只是不可能？

Answer 1

由于你有一个递归类型,你需要明确声明它:

data T a = L [a] | F (a -> T a)

(+>) (F f) = f
unL  (L x) = x

eat xs x = if x == "vomit" then L $ reverse xs else F $ eat (x:xs)

eaten = unL $ eat [] "x" +> "y" +> "z" +> "vomit"

*主要>吃了
["x","y","z"]

定义

eaten1 = eat [] "x" +> "y" 

eaten2 = eaten1 +> "z" +> "vomit"

也有效:

*Main> unL $ eaten1 +>"vomit"
["x","y"]
*Main> unL $ eaten2
["x","y","z"]