在Scheme中对函数进行n次咖喱 [英] Currying a function n times in Scheme

问题描述

我无法找出一种方法来咖喱函数指定次数.也就是说，我给该函数一个自然数n和一个有趣的函数，并且该函数将n次咖喱化.例如:

I'm having trouble figuring out a way to curry a function a specified number of times. That is, I give the function a natural number n and a function fun, and it curries the function n times. For example:

(curry n fun)

该函数是可能的应用程序吗?

Is the function and a possible application would be:

(((((curry 4 +) 1) 2) 3) 4)

哪个会产生10.

我真的不确定如何正确实施它.有人可以帮我吗?谢谢:)

I'm really not sure how to implement it properly. Could someone please give me a hand? Thanks :)

推荐答案

;;; no curry
;;; (Int, Int, Int) => Int

(define sum
  (lambda (x y z)
    (+ x y z) ) )

(sum 1 2 3)

;;; curry once
;;; (Int) => ((Int x Int) => Int)

(define sum
  (lambda (x)
    (lambda (y z)
      (+ x y z) ) ) )

(define sum+10 (sum 10))

(sum+10 10 20)

;;; curry 2 times
;;; (Int) => (Int => (Int => Int) )

(define sum
  (lambda (x)
    (lambda (y)
      (lambda (z)
        (+ x y z) ) ) ) )

(define sum+10+20 ((sum 10) 20))

(sum+10+20 30)

;;;现在，我们将从这些示例开始进行概括:

;;; Now we will generalize, starting from these examples:

(define (curry n f)
  (if (= n 0)
      (lambda (x) x)
      (lambda (x) (f ((curry (- n 1) f) x)))))

;;;示例:将函数1+的11倍应用于初始数字10，结果21:

;;;Example: apply 11 times the function 1+ on the initial number 10 , results 21:

((curry 11 (lambda (x) (+ x 1))) 10)

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