debugging,lisp,common-lisp , Common-LISP debugging (code included)

Common-LISP debugging (code included)


Tag: debugging,lisp,common-lisp

I hope to get a quick fix to my code, which takes in a list of numbers, numberlist, and a threshold and return the number of items in numberlist that are greater than threshold. I just can't figure out what's wrong, and I am not familiar with debugging. I am very new to stackoverflow, and LISP in general... any comments/criticism/advice will be welcome. Thank you!

ex) (count-greater-than (list 1 2 3 4 5 6 6 7) 5) => 3

(defun count-greater-than (numberlist threshold)
  (if (null numberlist) 0
    (counter (numberlist threshold 0)))
  (defun counter (numberlist threshold count)
    (cond ((null numberlist) count)
          ((> (first numberlist) threshold) (counter (rest numberlist) threshold (+ 1 count)))
          (t (counter (rest numberlist) threshold count)))))


First, note that the standard actually contains functions that will help with things like this. There's a useful count-if function that can take a predicate and count how many elements in a list satisfy it. For your case, you could do:

CL-USER> (count-if #'(lambda (x)
                       (> x 5))
                   (list 1 2 3 4 5 6 6 7))
;=> 3

CL-USER> (defun count-greater-than (numbers threshold)
           (count-if (lambda (n) (> n threshold)) numbers))
CL-USER> (count-greater-than (list 1 2 3 4 5 6 6 7) 5)
CL-USER> (count-greater-than (list 1 2 3 4 5 6 6 7) 6)

In your particular case, it looks like you're doing this manually, but you have the parenthesis wrong. It looks like you're trying to create a local helper function called counter. You can either define it outside the function with defun, e.g.:

(defun count-greater-than (numberlist threshold)
  (if (null numberlist) 0
    (counter (numberlist threshold 0))))

(defun counter (numberlist threshold count)
  (cond ((null numberlist) count)
        ((> (first numberlist) threshold) (counter (rest numberlist) threshold (+ 1 count)))
        (t (counter (rest numberlist) threshold count))))

or you could do it with a local definition, using labels:

(defun count-greater-than (numberlist threshold)
  (labels ((counter (numberlist threshold count)
             (cond ((null numberlist) count)
                   ((> (first numberlist) threshold) (counter (rest numberlist) threshold (+ 1 count)))
                   (t (counter (rest numberlist) threshold count)))))
    (if (null numberlist) 0
        (counter numberlist threshold 0))))


As Xach points out in the comments, you can actually do this even more succinctly using count's :test argument. I don't know whether it captures the notion of "count things with this property" quite as explicitly, but it makes for a very short solution:

CL-USER> (count 5 (list 1 2 3 4 5 6 6 7) :test #'<)
;=> 3

This counts how many times 5 is in the list, but the trick is that rather than checking whether a list element is 5 using eql or =, it uses <. That is, count will end up checking (< 5 1), then (< 5 2), …, up to (< 5 6), (< 5 7). It is specified that the test will be called with the arguments in that order. The glossary entry on satisfy the test says (emphasis added):

  1. (for a two argument test) to be in a state such that the two-place predicate which is the sequence function's test argument returns true when given a first argument that is the object being considered, and when given a second argument that is the result of calling the sequence function's key argument on an element of the sequence function's sequence argument which is being tested for equality;


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