方案：与集合混淆！
我很困惑这段代码是如何工作的：
(define m (list 1 2 3 '(5 8)))
(let ((l (cdr m)))
(set! l '(28 88))) ==>(1 2 3 (5 8))
(define o (list 1 2 3 '(5 8)))
(let ((l (cdr o)))
(set-car! l '(28 88))) ==> (1 (28 88) 3 (5 8))
为什么 (set! l '(28 88))) 不更新 m？
I'm confused how this code works:
(define m (list 1 2 3 '(5 8)))
(let ((l (cdr m)))
(set! l '(28 88))) ==>(1 2 3 (5 8))
(define o (list 1 2 3 '(5 8)))
(let ((l (cdr o)))
(set-car! l '(28 88))) ==> (1 (28 88) 3 (5 8))
Why does (set! l '(28 88))) not update m?
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评论（4）
放！不能改变列表（或任何其他结构，与 set-car/cdr 不同！），但只能更改 a （在您的情况下，本地l）变量的绑定。
set! can't mutate a list (or any other structure, unlike set-car/cdr!), but only change the binding of a (in your case, the local l) variable.
根据怀疑论者的回答， l 是一个本地绑定变量，并通过 set 更改它！不会对 m 做任何事。
如果您希望第一个 let 返回预期值，则需要通过返回本地绑定 l 的新值来进行后续操作，如下所示：
As per skeptic's answer, l was a locally-bound variable, and changing it via set! is not going to do anything to m.
If you want the first let to return the expected value, you need to follow-up by returning the new value of the locally-bound l, like so:
set!仅更改符号的绑定，通常让垃圾收集器消耗其原始值。它不会改变数据，而是重新绑定符号。因此，像(set! (car '(symbol symbol2)) 3)这样的东西不起作用，即使第二个子表单计算为符号值。要真正改变内存中的数据，可以使用
set-car!、set-cdr!、set-vector!等形式之一必须使用。它们具有完全不同的语义，并且确实评估其第二个子形式，并且无论其评估的数据是什么，都会在内存中更新，从而更改与其共享该内存的所有其他符号的值。set!only changes the binding of a symbol, often letting the garbage collector consume its original value. It doesn't mutate data, it rebinds a symbol. Hence things like(set! (car '(symbol symbol2)) 3)don't work, even though the second subform evaluates to a symbol-value.To truly mutate data in-memory, one of the forms of
set-car!,set-cdr!,set-vector!et cetera have to be used. These have completely different semantics and do evaluate their second sub-form, and what-ever data it evaluates to is then updated in-memory, changing all other symbol's value that share that memory with it.这与变异与重新绑定无关。
set!和set-car!都会更改指针。在您的示例中，l最初指向该对的cdr，它也是一对。当你设置时！ l，您告诉它指向保存'(28 88)的新位置。类似地，当您执行
set-car!时，您是在说car指针现在将指向位置 B 而不是其当前位置 A，但 A 的内容是没有变异。例如：It's not about mutating vs rebinding. Both
set!andset-car!change pointers. In your example,linitially points to thecdrof the pair, which is also a pair. When youset! l, you're telling it to point to the new location which holds'(28 88).Similarly, when you do
set-car!you are saying that thecarpointer will now point to a location B instead of its current location A, but the contents of A aren't mutated. For example: