CODEWITHSUNDEEP
scheme
bZhang
bZhang

Reputation: 307

scheme recursion - power of an int

I am trying to implement a function that tells if the parameter is power of 2. Here is what I have

   (define (powof2 x) (cond  
   [(and (even? x) (> x 1)) ((powof2 (/ x 2)))] 
   [else (equal?(x 1))]))

But when I try to run it with parameter 12 I get the error saying: Error: 3 is not a function [powof2, powof2, powof2, (anon)]

Any help?

Thanks!

Upvotes: 0

Views: 88

Answers (1)

uselpa
uselpa

Reputation: 18927

Ah those parenthesis ;-)

(define (powof2 x)
  (cond  
    [(and (even? x) (> x 1)) (powof2 (/ x 2))]
    [else (= x 1)]))

Note that in the 3rd line and 4th line you had an extra pair of parenthesis, and in the 4th line that you should use = to compare numbers.

Upvotes: 2

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