Reversing an integer - scheme

Question

I am trying to write a function which takes an input number and outputs the number in reverse order.
Ie:
Input -> 25
Output -> 52
Input -> 125
Output -> 521
I am new to lisp, if its helpful here is the working function in c++
function.cpp
int revs(int rev, int n)
{
if (n <= 0)
return rev;
return revs((rev * 10) + (n % 10), n/10);
}
I have written it in Racket as follows:
(define (revs rev n)
(if (<= n 0)
rev
(revs (+ (* rev 10) (modulo n 10)) (/ n 10))))
But when I run it with (revs 0 125) I get this error:
modulo: contract violation
expected: integer?
given: 25/2
argument position: 1st
other arguments...:
10
Certainly I am doing something incorrect here, but I am unsure of what I am missing.

Answer 1

The division operator / doesn't do integer division, but general division, so when you call, e.g., (/ 25 2), you don't get 12 or 13, but rather the rational 25/2. I think you'd want quotient instead, about which the documentation has:
procedure (quotient n m) → integer?
n : integer?
m : integer?
Returns (truncate (/ n m)). Examples:
> (quotient 10 3)
3
> (quotient -10.0 3)
-3.0
> (quotient +inf.0 3)
quotient: contract violation
expected: integer?
given: +inf.0
argument position: 1st
other arguments...:
3

Answer 2

Treating the operation lexicographically:
#lang racket
(define (lexicographic-reverse x)
(string->number
(list->string
(reverse
(string->list
(number->string x))))))
Works[1] for any of Racket's numerical types.
[edit 1] "Works," I realized, is context dependent and with a bit of testing shows the implicit assumptions of the operation. My naive lexicographic approach makes a mess of negative integers, e.g. (lexicographic-reverse -47) will produce an error.
However, getting an error rather than -74 might be better when if I am reversing numbers for lexicographic reasons rather than numerical ones because it illuminates the fact that the definition of "reversing a number" is arbitrary. The reverse of 47 could just as well be -74 as 74 because reversing is not a mathematical concept - even though it might remind me of XOR permutation.
How the sign is handled is by a particular reversing function is arbitrary.
#lang racket
;; Reversing a number retains the sign
(define (arbitrary1 x)
(define (f n)
(string->number
(list->string
(reverse
(string->list
(number->string n))))))
(if (>= x 0)
(f x)
(- (f (abs x)))))
;; Reversing a number reverses the sign
(define (arbitrary2 x)
(define (f n)
(string->number
(list->string
(reverse
(string->list
(number->string n))))))
(if (>= x 0)
(- (f x))
(f (abs x))))
The same considerations extend to Racket's other numerical type notations; decisions about reversing exact, inexact, complex, are likewise arbitrary - e.g. what is the reverse of IEEE +inf.0 or +nan.0?

Answer 3

Here is my solution for this problem
(define (reverseInt number)
(define (loop number reversedNumber)
(if (= number 0)
reversedNumber
(let ((lastDigit (modulo number 10)))
(loop (/ (- number lastDigit) 10) (+ (* reversedNumber 10) lastDigit)))))
(loop number 0))
Each time we multiply the reversed number by 10 and add the last digit of number.
I hope it makes sense.

Answer 4

A R6RS version (will work with R7RS with a little effort)
#!r6rs
(import (rnrs)
(srfi :8))
(define (numeric-reverse n)
(let loop ((acc 0) (n n))
(if (zero? n)
acc
(receive (q r) (div-and-mod n 10)
(loop (+ (* acc 10) r) q)))))
A Racket implementation:
#!racket
(require srfi/8)
(define (numeric-reverse n)
(let loop ((acc 0) (n n))
(if (zero? n)
acc
(receive (q r) (quotient/remainder n 10)
(loop (+ (* acc 10) r) q)))))

Answer 5

With recursion, you can do something like:
#lang racket
(define (reverse-num n)
(let f ([acc 0]
[n n])
(cond
[(zero? n) acc]
[else (f (+ (* acc 10) (modulo n 10)) (quotient n 10))])))