Question

(defn divisible [x y] (zero? (rem x y)))

1. ((or (fn [x] (divisible x 3)) (fn [x] (divisible x 5))) 3)
2. ((or (fn [x] (divisible x 3)) (fn [x] (divisible x 5))) 5)

the first expression evals to true but not the second one why?

• can some on explain whats going on here?

Answer 1

An expression of the form (or foo bar) does not “glue together” two predicates into one compound predicate; it returns foo if it is truthy, else bar. In your code, (fn [x] (divisible x 3)) is of course truthy (the only falsey values are false and nil), so the whole thing is equivalent to

((fn [x] (divisible x 3)) 3)
((fn [x] (divisible x 3)) 5)

What you want to do is something like

(some #(apply % 3) [(fn [x] (divisible x 3)) (fn [x] (divisible x 5)])
(some #(apply % 5) [(fn [x] (divisible x 3)) (fn [x] (divisible x 5)])
              ; ^- here goes the thing being tested

In general,

(defn or-preds [& preds]
  (fn [& args]
    (some #(apply % args) preds)))

((or-preds (fn [x] (divisible x 3)) (fn [x] (divisible x 5))) 3)
((or-preds (fn [x] (divisible x 3)) (fn [x] (divisible x 5))) 5)
;; or simpler...
((or-preds #(divisible % 3) #(divisible % 5)) 3)