Question

I have a function that takes a number such as 36, and reverses it to say

 '(6 3)

Is there anyway to combine that 6 3 to make it one number?

Here is the code that I have written.

    (define (number->rdigits num)
      (if (rdigits (/ (- num (mod num 10)) 10)))))


    (define reversible?
      (lambda (n)
        (cond
          [(null? n) #f]
          [else (odd? (+ n (list (number->rdigits n))))])))

Thanks!

Answer 1

You can do this using an iterative function that takes each element of the list in turn, accumulating a result. For example:

(define (make-number lst)
  (define (make a lst)
     (if (null? lst)
         a
         (make (+ (* 10 a) (car lst)) (cdr lst))))
  (make 0 lst))

(display (make-number '(6 3)))

The make function uses an accumulator a and the rest of the digits in lst to build up the final result one step at a time:

a = 0
a = 0*10 + 6 = 6
a = 6*10 + 3 = 63

If you had more digits in your list, this would continue:

a = 63*10 + 5 = 635
a = 635*10 + 9 = 6359

---

A less efficient implementation that uses a single function could be as follows:

(define (make-number lst)
  (if (null? lst)
      0
      (+ (* (expt 10 (length (cdr lst))) (car lst)) (make-number (cdr lst)))))

This function needs to calculate the length of the remainder of the list for each iteration, as well as calling the expt function repeatedly. Also, this implementation is not properly tail recursive so it builds up multiple stack frames during execution before unwinding them all after it reaches its maximum recursion depth.