Racket List Error Using Recursion

Question

I am trying to complete a school problem using lists in Racket. It is a simple problem using recursion but I cannot understand why my code won't work. We are supposed to search the list and return true if the item provided matches one in the list. This is what I have so far:

(define (containsAnywhere test  list)
  (cond 
    ((null? list) '())
    (equal?((car list) test))
    (else(containsAnywhere (test (cdr list))))))

But I get the following error:

application: not a procedure;
 expected a procedure that can be applied to arguments
  given: 1
  arguments.:

uselpa · Accepted Answer · 2016-02-06 23:24:43Z

A few comments:

don't call your parameter list, there's a build-in procedure of that name which will be shadowed this way; in Scheme it's common to call the parameter lst instead
you have a lot of parentheses errors - procedure calls are like (procedure param1 param2 ...), the conditions inside cond are in parentheses
the equal? predicate has a trailing ?
and your code's indentation is off; use Dr Racket's "reindent all" feature
I assume the result of your procedure is supposed to be True or False, so the empty list should return #f, not '()

Here's a working version of your code; inside cond I put the condition and the expression to be evaluated on separate lines so that the code and the parentheses become clearer:

(define (containsAnywhere test lst)
  (cond 
    ((null? lst)
     #f)
    ((equal? (car lst) test)
     #t)
    (else
     (containsAnywhere test (cdr lst)))))

Alternatively, I would code this like:

(define (containsAnywhere tst lst)
  (and (not (null? lst))
       (or (equal? (car lst) tst)
           (containsAnywhere tst (cdr lst)))))

sepp2k · Accepted Answer · 2016-02-06 21:51:49Z

0

(test (cdr list))

Here you're applying test to the argument (cdr list) as if test were a function, but it's not. Thus the error.

You presumably meant (containsAnywhere test (cdr list)), which passes test as the first argument to containsAnywhere and (cdr list) as the second.

answered Feb 6, 2016 at 21:51

sepp2k

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