方案中的连续传递风格? [英] Continuation-Passing Style in Scheme?

问题描述

我在维基百科上遇到了这段代码:

(define (pyth x y k)(* x x (拉姆达 (x2)(* y y (λ (y2)(+ x2 y2 (λ (x2py2)(sqrt x2py2 k)))))))))

文章说那段代码是另一段代码的Continuation-Passing版本:

(定义(pyth x y)(sqrt (+ (* x x) (* y y))))

然而，我很困惑:这到底是怎么工作的?在这里，你如何将一个数字乘以一个 lambda?(* x x (lambda ...))

解决方案

在维基百科示例中，* 与常规示例中的 * 的含义不同.

我会将维基百科的例子改写为:

(define (pyth x y k)(cps-* x x (拉姆达 (x2)(cps-* y y (λ (y2)(cps-+ x2 y2 (λ (x2py2)(cps-sqrt x2py2 k)))))))))

在这种形式中，每个 cps-xxx 函数执行指定的操作，然后将结果传递给最后一个参数.你可以这样称呼它:

(pyth 2 3 显示)

将 2 和 3 相乘，得到 6，然后将 6 传递给 display.(实际上，您可能希望将结果传递给 cps-display，后者显示其初始参数，然后调用指定为其最后一个参数的另一个函数).

I ran into this code on Wikipedia:

(define (pyth x y k)
    (* x x (lambda (x2)
        (* y y (lambda (y2)
            (+ x2 y2 (lambda (x2py2)
                (sqrt x2py2 k))))))))

The article says that that code is the Continuation-Passing version of another piece of code:

(define (pyth x y)
    (sqrt (+ (* x x) (* y y))))

However, I'm quite confused: How does that even work? How do you multiply a number by a lambda here? (* x x (lambda ...))

解决方案

In the Wikipedia example, * doesn't mean the same thing as * in the conventional example.

I would rewrite the Wikipedia example as:

(define (pyth x y k)
    (cps-* x x (lambda (x2)
        (cps-* y y (lambda (y2)
            (cps-+ x2 y2 (lambda (x2py2)
                (cps-sqrt x2py2 k))))))))

In this form, each of the cps-xxx functions perform the operation indicated and then pass the result to the last argument. You could call it like this:

(pyth 2 3 display)

which would multiply 2 and 3, giving 6, and then passing 6 to display. (Actually you would want to pass the result to a cps-display that displayed its initial argument(s) and then called another function specified as its last parameter).

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