使用汽车和cdr

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问题描述

我是新手，在使用car和cdr时遇到了困难.我在ast中有一个AST字符串文字.

I am new to scheme and having a hard time with using car and cdr. I have an AST string literal in ast.

(define ast '(program
  ((assign (var i int) (call (func getint void int) ()))
   (assign (var j int) (call (func getint void int) ()))
   (while (neq (var i int) (var j int))
    ((if (gt (var i int) (var j int))
         ((assign (var i int) (minus (var i int) (var j int))))
         ((assign (var j int) (minus (var j int) (var i int)))))))
   (call (func putint int void) ((var i int)))))
)

我知道汽车返回ast的头.所以

I know car returns the head of ast. So

(car ast)

返回程序".

我很困惑如何使用car和cdr从ast获取字符串，例如"assign"，"while"，"if"和"call".

I am confused how to use car and cdr to get strings from ast such as 'assign, 'while, 'if, and 'call.

推荐答案

您需要从球拍参考:

列表是递归定义的:它可以是常量null，或者是其第二个值是列表的一对.

A list is recursively defined: it is either the constant null, or it is a pair whose second value is a list.

基本上每个对(x . y)由两个元素组成-car让我们x cdr让我们y.

Basically every Pair (x . y) is made of two elements - car gets us x cdr gets us y.

请注意，x 和 y既可以成对也可以列出自己，就像您的AST一样，即(出于同一参考):

Notice that x and y can both be pairs or lists themselves, just like your AST, ie(out of same reference):

> (define lst1 (list 1 2 3 4))

>lst1 

'(1 2 3 4)

注意'(1 2 3 4)实际上是:(1 . ( 2 . ( 3 . ( 4 . ())))<-了解方案中的实现非常重要.

notice that '(1 2 3 4) is actually: (1 . ( 2 . ( 3 . ( 4 . ()))) <-- Very important to know the implementation in scheme.

> (car lst1)

1

> (cdr lst1)

'(2 3 4)

> (car (cdr lst1))

2

另一种链接汽车和cdr呼叫的方法(从右侧读取):cadr表示(cdr lst)，然后将car应用于答案=>. (car (cdr lst)) == (cadr lst)

Another way to chain car and cdr calls(read from the right):cadr means (cdr lst) and then apply car on the answer => (car (cdr lst)) == (cadr lst)

> (cdddr lst1)

'(4)

> (cadddr lst1)

4

> (define lst2 (list (list 1 2) (list 3 4)))

>lst2 

'((1 2) (3 4))

== ( ( 1 . ( 2 . ()) ) . ( 3 . ( 4 . () )))

> (car lst2) 

'(1 2)

>(cdr lst2)

'((3 4))

实际上是((3 . (4 . () ) ) . () ) == ((3 4) . ()) == ((3 4))

which is actually ((3 . (4 . () ) ) . () ) == ((3 4) . ()) == ((3 4))

您没有询问，但我假设您将遍历树/列表.最终，您将必须进行递归遍历(除非使用当前不适合的高级方法，即准备就绪时检查CPS)，就像这样:

You did not ask but I'm assuming you are going to traverse through the tree/list.Ultimately you will have to traverse with a recursion (unless using advanced method not suitable at this stage, ie check CPS when ready ) like so:

(define runner 
  (lambda (tree)
    (if (null? tree)
        null
        (let ((first_element (car tree))
              (rest_of_tree  (cdr tree)))
          ;body:
          ;do some logic like:
              ;if first is list call runner on it:
              ;(runner rest_of_tree)
              ;possibly chain answer of logic and call together
              ;else check/return string is there (recognize tree root)
          ))))

希望这有助于欢迎问题.

Hope this helps questions welcome.

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