Scheme中柯里化函数的实现

问题描述

当我执行以下操作时会发生什么?

What happens when I do the following?

(define ((func x) y)
    (if (zero? y)
        ((func x) 1)
        12))

我知道我可以做到这一点:

I understand that I can do this:

(define curried (func 5))

现在我可以使用咖喱了.我很好奇的是函数的定义.是否行

And now I can use curried. What I'm curious about is in the definition of the function. Does the line

((func x) 1)

以 x 作为参数创建一个新的 lambda，然后在 1 上调用它?或者它比那更聪明，它只是重新使用现有的.(例如，如果我执行 (curried 0)，((func x) 1) 行将等价于 (curried 1)- PLAI Scheme 会这样做吗?)

create a new lambda with x as the argument, and then invoke it on 1? Or is it smarter than that and it just re-uses the existing one. (For example, if I do (curried 0), the ((func x) 1) line would be equivalent to (curried 1) - does PLAI Scheme do this?)

推荐答案

Scheme 标准中规定

In the Scheme standard it is specified that

(define (f x) 42) is short for (define f (lambda (x) 42)) .

自然(非标准)概括意味着:

The natural (non-standard) generalization implies:

(define ((f x) y) (list x y)) is short for (define (f x) (lambda (y) (list x y)))
                which is short for (define f (lambda (x) (lambda (y) (list x y))))

为了测试它，让我们试试 DrScheme 中的例子

To test it, let's try the example in DrScheme

欢迎使用 DrScheme，版本 4.1.3.3-svn5dec2008 [3m].语言:模块；内存限制:384 MB.

Welcome to DrScheme, version 4.1.3.3-svn5dec2008 [3m].Language: Module; memory limit: 384 megabytes.

(define ((f x) y) (list x y))(f 1)

((f 1) 2)(1 2)

((f 1) 2) (1 2)

如果我们命名临时值，可能更容易看到发生了什么:

If we name the temporary value, it might be easier to see what happens:

(定义 h (f 1))(h 2)(1 2)(h 3)(1 3)

由于PLAI Scheme"是在 DrScheme 中实现的，我相信它继承了这个快捷方式.

Since "PLAI Scheme" is implemented in DrScheme, I believe it inherits this shortcut notation.

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