可以“延伸”吗？ Scheme中的函数/ lambda / macro？ [英] Is it possible to "extend" a function / lambda / macro in Scheme?

问题描述

例如：如果我希望函数相等？识别我自己的类型或记录，我可以添加的新行为吗？？没有删除或覆盖旧的？

For example: if I want the function equal? recognize my own type or record, can I add a new behavior of equal?? without erasing or overwriting the old one?

或者例如，如果我想使函数+接受还有字符串？

Or for example if I want to make the function "+" accept also string?

推荐答案

不是使用 import ，更好的解决方案是通过让 - 绑定它来跟踪原始功能。最好还是检查参数的类型是否为字符串，而不是它不是数字。使用这两种方法意味着可以组合技术。

Rather than using import, a better solution is to keep track of the original function by let-binding it. It's also better to check that the type of the argument is a string, rather than that it is not a number. Using both of these approaches means that it's possible to compose the technique.

(define +
  (let ((old+ +))
    (lambda args
      (if (string? (car args))
          (apply string-append args)
          (apply old+ args)))))

(define +
  (let ((old+ +))
    (lambda args
      (if (vector? (car args))
          (apply vector-append args)
          (apply old+ args)))))

以上将产生一个 + 适用于数字，字符串或向量的函数。一般来说，这是一种更具扩展性的方法。

The above will produce a + function that works on numbers, strings, or vectors. In general, this is a more extensible approach.

我能够在MIT / GNU中验证上述方法是否正常工作Scheme，Guile，Racket，Chicken，TinyScheme和SCSH。但是，在某些实现中，例如Biwa Scheme，有必要使用 set！而不是 define 。在Ikarus中， set！不能用于导入的基元，而 define 会破坏环境，所以有必要分两步执行此操作：

I was able to verify that the above works correctly in MIT/GNU Scheme, Guile, Racket, Chicken, TinyScheme, and SCSH. However, in some implementations, such as Biwa Scheme, it is necessary to use set! instead of define. In Ikarus, set! cannot be used on an imported primitive, and define messes up the environment, so it is necessary to do this in two steps:

(define new+
  (let ((old+ +))
    (lambda args
      (if (string? (car args))
          (apply string-append args)
          (apply old+ args)))))
(define + new+)

---

请注意，根据 R5RS ， define 和 set！在这种情况下应该是等价的：

---

Note that according to R5RS, define and set! are supposed to be equivalent in this case:

在程序的顶层，定义

At the top level of a program, a definition

(define <variable> <expression>)

与赋值表达式具有基本相同的效果

has essentially the same effect as the assignment expression

(set! <variable> <expression>)

if < variable> 已绑定。

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