you don't need 'fn' when you're using #(...) - that is the whole point.
To not have to declare a function and its args explicitly.
this particular example you cannot write using #() syntax though because
it is recursing at 2 points and you cannot use 'recur'.
Jim
wOn 13/11/13 17:41, Angus wrote:
I know this fibonacci function is not optimal but I want to learn one
step at a time. I want to be able to apply this approach to 4clojure
which disallows a lot of things including defn.
This is my implementation so far:
(defn fib [x]
(cond
(= x 0) 0
(= x 1) 1
:else
(+ (fib (- x 1)) (fib (- x 2)))))
So I thought about something like this:
(#(fn fib [x]
(cond
(= x 0) 0
(= x 1) 1
:else
(+ (fib (- x 1)) (fib (- x 2)))))
8)
But trying that in a REPL I get error:
clojure.lang.ArityException: Wrong number of args (1) passed to:
sandbox28956$eval28971$fn
So how can I call this as a one line REPL?
I though using apply with a one element list might work, but no.
(apply #(fn fib [x](cond (= x 0) 0 (= x 1) 1 :else (+ (fib (- x 1))
(fib (- x 2))))) '(8))
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