CODEWITHSUNDEEP
if-statementscheme
Ofek Ron
Ofek Ron

Reputation: 8580

Not a Procedure, Really?

I been trying to figure this out for like 2 hours now! please Consider the following code:

(define (PowListF list) 
  (PowListFHelp list (- (length list) 1)))

(define (FuncPow f n) 
  (if (= 0 n)
      f
      (lambda (x)
        (FuncPow (f (- n 1)) x))))

(define (PowListFHelp list n) 
  (if (= n 0)
      (list (list-ref list 0))
      (cons (FuncPow (list-ref list n) n)
            (PowListFHelp list (- n 1)))))

The rocket Scheme compilers give me error: application: not a procedure; expected a procedure that can be applied to arguments given: (# #) arguments...: #

and it points to the (cons part...

and The following code, that uses the same if structure, does work:

(define (polyCheb n)
  (reverse (polyChebHelp n)))

(define (polyChebHelp n)
  (if (= n 0)
      (list (polyChebFunc n))
      (cons (polyChebFunc n)
            (polyChebHelp (- n 1)))))

(define (polyChebFunc n)
  (if (= n 0)
      (lambda (x) 1)
      (if (= n 1) 
          (lambda (x) x)     
          (lambda (x)
            (- (* (* 2 x)
                  ((polyChebFunc(- n 1)) x))
               ((polyChebFunc(- n 2)) x))))))

EDIT: PowListF is being given list of functions as parameter, returns the same list s.t every function is composed with itself (its index + 1) times...

usage: ((list-ref (PowListF (list (lambda (x) x) (lambda (x) (expt x 2)))) 1) 2) value should be 2^2^2=2^4=16

EDIT 2:

This was the solution : 

(define (FuncPow f n) 
  (if (= 0 n)
      f
      (lambda (x) (f ((FuncPow f (- n 1)) x)))))

Upvotes: 1

Views: 121

Answers (1)

leppie
leppie

Reputation: 117220

The problem lies with: (FuncPow(f (- n 1)) x)

The result of calling f must be procedure (I assume this as you return a procedure).

You further then 'assign' f as: (list-ref list n).

Upvotes: 2

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