How to convert this simple (and inefficient) fibonacci function to # format

[Angus]

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))

[Ben Wolfson]

#(fn fib [x] ... ) creates a zero-arity function which, when called, will return a one-arity function. Just get rid of the #.

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--
Ben Wolfson
"Human kind has used its intelligence to vary the flavour of drinks, which may be sweet, aromatic, fermented or spirit-based. ... Family and social life also offer numerous other occasions to consume drinks for pleasure." [Larousse, "Drink" entry]

[Jim - FooBar();]

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

[Angus]

Ah, as in:

> ((fn fib [x](cond (= x 0) 0 (= x 1) 1  :else  (+ (fib (- x 1)) (fib (- x 2))))) 8))

21

Many thanks.

[angus...@gmail.com]

Did this get returned from my htc

Sent from my HTC

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