LISP程序输出

Question

我是LISP的新手，无法弄清楚以下LISP会做什么？

(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) 

第一行分配3个元素分配给符号A的列表。引用这是为了防止将(RIGHT ARE YOU)作为RIGHT的函数进行评估。第二行做了一些不必要的冗长和复杂的逻辑，基本上创建了一个由四个元素组成的列表：字符串HOW和A中的三个元素。

打破的第二行：

• (first (rest A)) - 这从A
• (first (rest (rest A)))返回元素YOU - 这从A
• (first A)返回元素ARE - 此返回从元件 'RIGHT' A

Th现在留给你：

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

哪个LISP是你学习？许多LISP都有一个REPL的概念（read-eval-print loop），它允许您试验复杂的表达式并将它们分解为更小的块，以了解中间步骤的结果。

Answer 2

也许这份成绩单用Common Lisp REPL会议将启发：

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>