removing an item from a list in scheme

问题:

How would you update a list by removing the first item?

consider the following pseudocode

define remover
(list = cdr list))

or is there a more recommendable way of removing the first item in a list?

回答1:

You nailed it, just write it as a procedure:

(define (remover lst)
  (cdr l))

And use it like this (this creates a new binding, and is not an assignment):

(let ((new-list (remover old-list)))
  new-list)

Or like this (this defines a new list):

(define new-list (remover old-list))

In any case, be aware that the original list passed as a parameter to remover will not be modified, instead a new list will be returned without the first element of the old list; that's the way to operate on immutable linked lists, you should never assume that any modifications on a list will occur in-place.

回答2:

Beware!

There are no list data structure per se. So you can not take a list, l, and remove the first element of l.

What you can do, when given a list l, is to use cdr to produce a new list which has the same elements as l does, except that the new list doesn't hold the first element.

A little more detail: A list holding the three values 1, 2, and, 3 is represented as (cons 1 (cons 2 (cons 3 '())). Let us name the cons-cells:

       c3 = (cons 3 '()) 
       c2 = (cons 2 c3)
   l = c1 = (cons 1 c2)

The first thing to note is that the entire list is given by the value c1. We can not delete the number 1 from the list by manipulating the cons-cell c1. We can however easily find a list omitting the first element, since

   c2 = (cons 2 c3) = (cons 2 (cons 3 '())

Therefore (cdr l) = (cdr c1) = c2 will produce a new list omitting the first element.

回答3:

Think about what cdr does. Remember, cdr returns a list. Your answer is close, you just need to think about it in terms of Scheme syntax and functions.