LISP程序输出

Question

我对LISP很陌生，不知道下面的LISP是做什么的？

(setq A '(RIGHT ARE YOU))
(print (reverse (list (first (rest A))(first (rest (rest A))) (first A) 'HOW)))

setq分配词法变量

Answer 1

它打印：

(HOW RIGHT YOU ARE)

第一行将3个元素的列表分配给符号A。这是为了防止将(RIGHT ARE YOU)计算为称为RIGHT的函数。第二行执行一些不必要的冗长和复杂的逻辑，基本上创建一个由四个元素组成的列表:字符串HOW和来自A的三个元素。

打破第二条线：

• (first (rest A)) -这将从A返回元素YOU
• (first (rest (rest A))) -这将从A返回元素ARE
• (first A) -这将从A返回元素“右”

这现在留给你的是：

(print (reverse (list ARE YOU RIGHT 'HOW)))

你在学哪种口音？许多LISP有一个REPL (读- experiment循环)的概念，它允许您对复杂表达式进行实验，并将它们分解成较小的块，以了解中间步骤的结果。

Answer 2

也许这个REPL会议的记录将启发我们：

CL-USER> (setq a '(right are you))
(RIGHT ARE YOU)
CL-USER> (print (reverse (list (first (rest a)) (first (rest (rest a))) (first a) 'how)))

(HOW RIGHT YOU ARE) 
(HOW RIGHT YOU ARE)
CL-USER> a
(RIGHT ARE YOU)
CL-USER> (rest a)
(ARE YOU)
CL-USER> (cdr a)
(ARE YOU)
CL-USER> (first (rest a))
ARE
CL-USER> (cadr a)
ARE
CL-USER> (rest (rest a))
(YOU)
CL-USER> (cddr a)
(YOU)
CL-USER> (first (rest (rest a)))
YOU
CL-USER> (caddr a)
YOU
CL-USER> (first a)
RIGHT
CL-USER> (car a)
RIGHT
CL-USER> (values (first (rest a)) (first (rest (rest a))) (first a) 'how)
ARE
YOU
RIGHT
HOW
CL-USER> (list (first (rest a)) (first (rest (rest a))) (first a) 'how)
(ARE YOU RIGHT HOW)
CL-USER> (list 'are 'you 'right 'how)
(ARE YOU RIGHT HOW)
CL-USER> (reverse '(are you right how))
(HOW RIGHT YOU ARE)
CL-USER> '(how right you are)
(HOW RIGHT YOU ARE)
CL-USER> (print '(how right you are))

(HOW RIGHT YOU ARE) 
(HOW RIGHT YOU ARE)
CL-USER>