使用汽车和cdr [英] Using car and cdr

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

一对精确地结合了两个值.第一个值是通过car过程访问的，第二个值是通过cdr过程访问的.对不是可变的(但请参见可变对和列表).

A pair combines exactly two values. The first value is accessed with the car procedure, and the second value is accessed with the cdr procedure. Pairs are not mutable (but see Mutable Pairs and Lists).

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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