常见Lisp中的指针 [英] Pointers in Common Lisp

问题描述

我想保存对我保存在另一个变量中的某些数据的一部分的引用(指针):

I want to save a reference (pointer) to a part of some Data I saved in another variable:

(let ((a (list 1 2 3)))
  (let ((b (car (cdr a)))) ;here I want to set b to 2, but it is set to a copy of 2
    (setf b 4))
  a) ;evaluates to (1 2 3) instead of (1 4 2)

我可以使用宏，但是如果我想在列表中间更改某些数据，那么我将需要执行很多代码，而且我也不是很灵活:

I could use macros, but then there would ever be much code to be executed if I want to change some Data in the middle of a list and I am not very flexible:

(defparameter *list* (create-some-list-of-arrays))
(macrolet ((a () '(nth 1000 *list*)))
  (macrolet ((b () `(aref 100 ,(a))))
    ;; I would like to change the macro a here if it were possible
    ;; but then b would mean something different
    (setf (b) "Hello")))

是否可以将变量创建为引用而不是副本?

Is it possible, to create a variable as a reference and not as a copy?

推荐答案

cl-user> (let ((a '(1 2 3)))
           (let ((b (car (cdr a))))
             (setf b 4))
           a)
;Compiler warnings :
;   In an anonymous lambda form: Unused lexical variable B
(1 2 3)

cons单元格是一对指针. car取消引用第一个，而cdr取消引用第二个.您的列表实际上是

A cons cell is a pair of pointers. car dereferences the first, and cdr dereferences the second. Your list is effectively

  a -> [ | ] -> [ | ] -> [ | ] -> NIL
        |        |        |
        1        2        3

在您定义b的顶部，(cdr a)使您获得第二个箭头.取其中的car会取消引用该第二个单元格的第一个指针，并将其值交给您.在这种情况下，为2.如果要更改该指针的值，则需要setf而不是其值.

Up top where you're defining b, (cdr a) gets you that second arrow. Taking the car of that dereferences the first pointer of that second cell and hands you its value. In this case, 2. If you want to change the value of that pointer, you need to setf it rather than its value.

cl-user> (let ((a '(1 2 3)))
           (let ((b (cdr a)))
             (setf (car b) 4))
           a)
(1 4 3)

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