Question

I have this Clojure code:

(defn apply-all-to-arg [& s]
    (let [arg (first s)
          exprs (rest s)]
        (for [condition exprs] (condition arg))))

(defn true-to-all? [& s]
    (every? true? (apply-all-to-arg s)))

This is test code:

(apply-all-to-arg 2 integer? odd? even?)

=> (true false true)

(every? true? (apply-all-to-arg 2 integer? odd? even?)

=> false

(true-to-all? 2 integer? odd? even?)

=> true

My question is:
Why does the function true-to-all? return true (it must have returned false instead)

Answer 1

true-to-all? calls apply-all-to-arg with the single argument s. So you’re not calling (every? true? (apply-all-to-arg 2 integer? odd? even?), but rather:

(every? true? (apply-all-to-arg (list 2 integer? odd? even?))

So in apply-all-to-arg the value of arg will be that list and the value of exprs will be the empty list. Since every? will be true for the empty list no matter what the condition is, you’ll get back true.

To fix this you can either change apply-all-to-arg, so that it accepts a list instead of a variable number of arguments, or you can change true-to-all?, so that it passes the contents of s as multiple arguments rather than a single list (by using apply).