Are clojure function cyclic dependencies specifically disallowed by design, or is it just a reader behaviour?

Question

If I do the following in clojure

(defn sub1a [a]
  (cond
    (= a 0) 0
    true (sub1b (- a 1) )))

(defn sub1b [a]
  (cond
    (= a 0) 0
    true (sub1a (- a 1) )))

(println (sub1a 10))

I get the following error:

java.lang.Exception: Unable to resolve symbol: sub1b in this context

But if I do the following:

(defn sub1a [a]
  (cond
    (= a 0) 0
    true (- a 1)))

(defn sub1b [a]
  (cond
    (= a 0) 0
    true (- a 1)))

(defn sub1a [a]
  (cond
    (= a 0) 0
    true (sub1b (- a 1) )))

(defn sub1b [a]
  (cond
    (= a 0) 0
    true (sub1a (- a 1) )))

(println (sub1a 10))

It runs just fine.

Is this by design, or just a function of the way the Clojure reader works?

Answer 1

You can do

(declare sub1a sub1b)

'declare' is specifically meant to create a var with no bindings for making forward declarations.

One you declared the names:

(defn sub1a [a]
  (cond
    (= a 0) 0
    true (sub1b (- a 1) )))

(defn sub1b [a]
  (cond
    (= a 0) 0
    true (sub1a (- a 1) )))

(println (sub1a 10))

Also the idomatic way to specify the default condition in cond (for clojure) is using the :else clause. This is a bit different from Common Lisp which uses T (for True). So your previous code can be rewritten as:

(defn sub1a [a]
  (cond
    (= a 0) 0
    :else (sub1b (- a 1) )))

...