Question

I kinda understand how to convert elementary functions such as arithmetics to continuation-passing style in Scheme.
But what if the function involves recursion?
For example,

(define funname 
     (lambda (arg0 arg1)
                (and (some procedure)
                      (funname (- arg0 1) arg1))))

Please give me advices.
Thank you in advance.

Answer 1

(define (func x y k)
  (some-procedure
   (lambda (ret)
     (if ret
         (- x 1
            (lambda (ret)
              (func ret y k)))
         (k #f))))

You are lacking a base case, which is why the only explicit call to the continuation is (k #f). If you have a base case, then you’d pass the base case return value to the continuation, also. For example:

(define (func x y k)
  (zero? x
         (lambda (ret)
           (if ret
               (k y)
               (some-procedure
                (lambda (ret)
                  (if ret
                      (- x 1
                         (lambda (ret)
                           (func ret y k)))
                      (k #f))))))))