I have a Clojure function:
(def obseq
(fn []
(let [a 0
b 1
c 2]
(println (seq '(a b c))))))
It outputs :
(a b c)
I want it to output a sequence containing the values of a, b, and c, not their names.
Desired output:
(1 2 3)
How do I implement this?
clj.core=> (defn obseq []
(let [a 0
b 1
c 2]
(println [a b c])))
#'clj.core/obseq
clj.core=> (obseq)
[0 1 2]
nil
Quoting a form like '(a b c) recursively prevents any evaluation inside the quoted form. So, the values for your 3 symbols a, b, and c aren't substituted. It is much easier to use a vector (square brackets), which never needs to be quoted (unlike a list). Since the vector is unquoted, the 3 symbols are evaluated and replaced by their values.
If you wanted it to stay a list for some reason, the easiest way is to type:
clj.core=> (defn obseq [] (let [a 0 b 1 c 2] (println (list a b c))))
#'clj.core/obseq
clj.core=> (obseq)
(0 1 2)
nil
This version also has no quoting, so the 3 symbols are replaced with their values. Then the function (list ...) puts them into a list data structure.
Note that I also converted your (def obseq (fn [] ...)) into the preferred form (defn obseq [] ...), which has the identical result but is shorter and (usually) clearer.