在 Scheme 或 Racket 中检测函数的调用者 [英] Detecting the caller of a function in Scheme or Racket

问题描述

在 Scheme 或 Racket 中是否可以检测函数的调用者?

In Scheme or Racket is it possible to detect the caller of a function?

例如，我可以编写一个函数来测试一个列表是否是一个原子列表，如下所示:

For example, I can write a function to test if a list is a list of atoms as follows:

(define atom? (lambda (x) (and (not (pair? x)) (not (empty? x)))))

(define lat? (lambda (l)
               (define latt?
                 (lambda (l)
                   (cond
                     ((null? l) #t)
                     ((atom? (car l)) (latt? (cdr l)))
                     (else #f))))
               (if (null? l) #f (latt? l))))

但不是上面的，有没有像调用者"这样的函数来做这样的事情:

but instead of the above, is there a function like "called-by" to do something like this:

(define lat?
  (lambda (l)
    (cond
      ((and (null? l) (called-by "lat?")) #t)
      ((atom? (car l)) (lat? (cdr l)))
      (else #f))))

推荐答案

通常的方法是向函数添加一些参数，或者像您一样通过内部定义进行循环.除此之外，没有可靠的方法来找出函数的调用者.

The usual way to do this is to add some argument to the function, or make a loop via an internal definition as you did. Other than that, there is no reliable way to find out the caller of a function.

但在您的情况下，似乎缺少功能 - 将其用于上述问题非常糟糕.内部助手版本没有任何问题.(它也与任何其他语言非常相似.)

But in your case, it seems like a good lack of feature -- using it for the above problem is pretty bad. There's nothing wrong with the internal helper version. (It's also quite similar to any other language.)

最后，我希望 (lat?null) 返回 #t 因为它是一个只有原子作为元素的列表.

Finally, I'd expect (lat? null) to return #t since it is a list that has only atoms as elements.

这篇关于在 Scheme 或 Racket 中检测函数的调用者的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！