更新整个结构 [英] Update the whole structure

问题描述

假设我有一些返回结构的函数:

Suppose I have some function which returns a struct:

(struct layer (points lines areas))

(define (build-new-layer height)
  ...
  (layer list-a list-b list-c))

我想跟踪最后返回的结果，例如:

I want to keep track of the last returned result something like:

(define top-side (build-new-layer 0)) ; store the first result
...
(set! top-side (build-new-layer 0.5)) ; throw away the first result and store the new one

但是，对于该特定代码，我得到了错误:

However, for that particular code I get the error:

set!: assignment disallowed;
 cannot modify a constant
  constant: top-side

请告诉我什么是做我想要的正确方法

Please, tell me what would be the right way to do what I want

推荐答案

您使用的是哪种语言?看来这是一个配置问题，因为从原则上来说，您所做的应该可以工作.转到选择语言"窗口(在Windows中为Ctrl + L)，单击显示详细信息"，然后查看当前使用的语言的选项之一是否不允许重新定义变量.或者，尝试使用其他语言.

What language are you using? it seems it's a matter of configuration, because in principle what you're doing should work. Go to the "choose language" window (Ctrl+L in Windows), click on "show details" and see if one of the options of the language currently in use disallows redefinition of variables. Alternatively, try using a different language.

根据 where 的确切含义，您将要使用存储的结果(我无法从问题代码中看出)，您可以通过这种方式将其作为函数参数传递不再需要使用全局变量.在Scheme中不建议依赖全局状态和突变(set!操作)，这可能是一个更好的主意.

Depending on where exactly you're going to use the stored result (I can't tell from the code in the question), you could pass it around as function parameters, in such a way that using a global variable is no longer necessary. This might be a better idea, relying on global state and mutation (the set! operation) is discouraged in Scheme.

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