The goal is to eliminate all inner parenthesis. 

> 
> (flatten '(a (b c) d)) becomes '(a b c d)
> 
> 
> 

This is my code in Racket

```
; if slist is null, return empty
; otherwise, if it is a pair, recursively solve car and cdr and concat them
; if it is a symbol, return the symbol

(define flatten
  (lambda (slist)
    (cond
      [ (null? slist) '()]
      [ (pair? slist)  
        (cons ((flatten (car slist)) (flatten (cdr slist))))]
      [ (symbol? slist) slist])))

```

It's complaining 

```
procedure application: expected procedure, given: c; arguments were: ()

```

which means I am trying to access `car` and `cdr` of an empty list.

```
I did the trace:
> (flatten '(a (b c) d))
pair?-car-cdr
a
((b c) d)
symbol?
a
pair?-car-cdr
(b c)
(d)
pair?-car-cdr
b
(c)
symbol?
b
pair?-car-cdr
c
()
symbol?
c
(stops here)

```

The trace code is simple - a bunch of displays.

```
(define flatten
  (lambda (slist)
    (cond
      [ (null? slist) '()]
      [ (pair? slist) 
        (display 'pair?-car-cdr)
        (newline)
        (display (car slist))
        (newline)
        (display (cdr slist))
        (newline)
        (cons ((flatten (car slist)) (flatten (cdr slist))))]
      [ (symbol? slist) 
         (display 'symbol?)
         (newline)
         (display slist)
         (newline)
        slist])))

```

What I don't understand is how come the first condition `(null? slist)` didn't catch the empty list? I have two recursive calls. If it did catch the empty list, it would go to the next recursion which is the list `{d}`.

What is the problem with my recursion logic?

---

Update version

```
(define flatten
  (lambda (slist)
    (cond
      [ (null? slist) '()]
      [ (pair? slist)  
        (cons (flatten (car slist)) (flatten (cdr slist)))]
      [ (symbol? slist) slist])))

(display (equal? (flatten '(a (b a) b a c (a b) c (e f (b a)))) '(a b a b a c a b c e f b a)))
(newline)
(display (equal? (flatten '(a b c)) '(a b c)))
(newline)
(display (equal? (flatten '(a (b c))) '(a b c)))
(newline)
(display (equal? (flatten '((a)(b)(c) d)) '(a b c d)))
(newline)
(display (equal? (flatten '(a (b) ((c)) (((d))) ((((e (f g))))))) '(a b c d e f g )))
(newline)
(display (equal? (flatten '()) '()))
(newline)
(display (equal? (flatten '(a b () ())) '(a b)))
(newline)

```

As Ross Larson suggested, append will make the program works. But for the sake of learning, if any one have the time to spare, the result of my tests show only the base cases passed (2nd and the empty list)

I thought about writing a wrapper function which calls `(cons (flatten slist) empty)`