Scheme append procedure

I'm having trouble 开发者_运维技巧appending a list to another list. Below is my code. When I run (append '(1 2) '(3 4)) I get '(1 3 2 4).

I want the output to be '(1 2 3 4)

(define (append l m)
 (if (null? l) '()
  (cons (car l) (append m (cdr l)))))

Thanks

Well by switching the two lists around like that (switching the position of m and l when calling append recursively), you'll get the first item of the first list followed by the first item of the second list etc.

If you don't want that, you should keep l as the first argument and m as the second. So you get:

(define (append l m)
 (if (null? l) '()
  (cons (car l) (append (cdr l) m))))

Of course this doesn't work as wanted either, because now you only get the first list back and nothing is appended at all. What you need to do is, once the first list is fully appended (i.e. once l is empty), you need to return the second one as the tail, like this:

(define (append l m)
 (if (null? l) m
  (cons (car l) (append (cdr l) m))))

I came across this while studying myself. @sepp2k's answer is a fine piece of instruction guiding OP to correct their code to achieve a recursive definition of append. Here's an alternate definition of my-append using the higher-order function foldr:

(define (myappend xs ys)
  (foldr cons ys xs))