AND / OR可以使用递归函数来获取值并在Common Lisp中使用它吗?

时间:2012-03-15 20:50:24

标签: function recursion common-lisp

我被困在这一段时间了。我试过了,我没有得到它。我以为我得到了它,但事实上这不起作用令人困惑。我应该得到1,但一直都是零。目的是使用规则简化表达式(我在下面添加)。我的问题:

(defun simplify (main-list)
  (setq count 1)

  (if (and (eq t (atom (car (cdr main-list))))
           (eq t (atom (car (cdr (cdr main-list))))))
      (print "this says that the 2 ands are cons cells"))

  (if (and (eq nil (cdr main-list))
           (eq t (atom (car main-list))))
      (print "reached end of file, going back now"))

  (if (eq 'and (car main-list))
      (progn
        (if (and (eq t (atom (car (cdr main-list))))
                 (eq nil (atom (car (cdr (cdr main-list))))))
            (if (or (eq nil (car (cdr main-list)))
                    (simplify (car (cdr (cdr main-list)))))
                nil
                (if (eq 1 (car (cdr main-list)))
                    (simplify (car (cdr (cdr main-list))))
                    (if (eq 1 (simplify (car (cdr (cdr main-list)))))))))

        (if (and (eq t (atom (car (cdr main-list))))
                 (eq t (atom (car (cdr (cdr main-list))))))
            (if (or (eq nil (car (cdr main-list)))
                    (eq nil (car (cdr (cdr main-list)))))
                nil
                (if (eq 1 (car (cdr main-list)))
                    (car (cdr (cdr main-list)))
                    (if (eq 1 (car (cdr (cdr main-list))))
                        (car (cdr main-list)))))))))

我使用的列表是:

(and 1 (and 1 1))

这是我想要完成的一个简单版本,但我正在逐步解决这个问题,因为我完全不熟悉该语言。这些是我想要遵循的家庭作业的规则:

(and x nil) => nil; 
(and nil x) => nil;
(and x 1) => x; 
(and 1 x) => x;

我已经通过

测试了它
(simplify (car(cdr(cdr x))))

我添加了计数以查看它是否均匀循环,但事实并非如此。所以我的猜测它与第一个代码块中if语句中的递归函数调用有关。对于为什么会有任何解释将不胜感激。

1 个答案:

答案 0 :(得分:0)

(defun simplify (expression)
  (if (consp expression)
      (destructuring-bind (op arg1 arg2)
          expression
        (ecase op
          (and (cond ((or (null arg1) (null arg2))
                      nil)
                     ((eql 1 arg1)
                      (simplify arg2))
                     ((eql 1 arg2)
                      (simplify arg1))))))
    expression))