Push 不会修改作为函数参数的列表 [英] Push doesn't modify the list being a function argument
问题描述
我是普通 lisp 的新手,所以希望有人能向我澄清这一点:
I'm new to common lisp, so hope someone would clarify this to me:
假设我们有一个列表,想用 push
添加一个项目来修改它:
say we have a list and want to add an item with push
to modify it:
CL-USER> (defparameter xx '(1 2 3))
XX
CL-USER> xx
(1 2 3)
CL-USER> (push 100 xx)
(100 1 2 3)
CL-USER> xx
(100 1 2 3)
正如预期的那样.但是当我尝试对函数执行相同操作时,它不会修改列表:
as expected. But when i try to do the same with the function, it doesn't modify a list:
CL-USER> (defun push-200 (my-list)
(push 200 my-list))
PUSH-200
CL-USER> (push-200 xx)
(200 100 1 2 3)
CL-USER> xx
(100 1 2 3)
所以我尝试像这样比较参数和我的列表:
so i tried to compare argument and my list like this:
CL-USER> (defun push-200 (my-list)
(format t "~a" (eq my-list xx))
(push 200 my-list))
WARNING: redefining COMMON-LISP-USER::PUSH-200 in DEFUN
PUSH-200
CL-USER> (push-200 xx)
T
(200 100 1 2 3)
CL-USER> xx
(100 1 2 3)
它说对象是相同的.所以问题是:我在这里忽略了什么?
it says the objects are identical. So the question is: what was the thing I've overlooked here?
推荐答案
(defun push-200 (my-list)
(push 200 my-list))
这会修改变量 my-list
.变量现在指向一个新的cons.
This modifies the variable my-list
. The variable now points to a new cons.
基本上是:(setq my-list (cons 200 my-list)
.
(push-200 xx)
Common Lisp 将 xx
计算为一个值并将列表传递给 push-200
.xx
不是传递的,而是它的值.因此,Lisp 系统不能修改 xx
以指向不同的 cons,因为它没有被传递.
Common Lisp evaluates xx
to a value and passes the list to push-200
. xx
is not passed, but the value of it. So, the Lisp system can't modify xx
to point to a different cons, since it is not passed.
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