Subj : Re: Reversing a number
To   : comp.programming
From : pete
Date : Tue Sep 13 2005 01:02 pm

Rob Thorpe wrote:
> 
> Willem wrote:
> > Roger wrote:
> > ) "Irrwahn Grausewitz" <irrwahn35@freenet.de> wrote in message
> > ) news:shv7i1p3on0qpsr2rumeifst69o8tjtp32@4ax.com...
> > )> But why treat the first iteration as a special case?
> > )> Just replace the two lines above with:
> > )>
> > )>     int n = 0;
> > )>
> > )>>    while (i)
> > )>>    {
> > )>>        n *= 10;
> > )>>        n += i % 10;
> > )>>        i /= 10;
> > )>>    }
> > )>>    return n;
> > )>>}
> > )
> > ) Then reverse(10000) is (1).
> >
> > Of course it is; the function returns an integer, after all.
> 
> I think the OP means a number in the mathematical sense of the word.
> ie:-
> 
> 1 -> 1
> 21 -> 12
> 5634 -> 4365
> 
> As Jerry Quinn pointed out 
> it's simplest to convert it to a string then
> reverse it and convert it back again.

#include <stdlib.h>
#include <limits.h>

int revdecimal(int number);
void itoa_r(int n, char *s);

int revdecimal(int number)
{
    char string[sizeof number * CHAR_BIT / 3 + 1];
    
    itoa_r(number, string);
    return atoi(string);
}

void itoa_r(int n, char *s)
{
     int tenth, min_offset;
     char *p;
     
     min_offset = 0;
     if (0 > n) {
         if (-INT_MAX > n) {
             ++n;
             ++min_offset;
         }
         n = -n;
         *s++ = '-';
     }
     p = s;
     tenth = n;
     do {
         tenth /= 10;
         *p++ = (char)(n - 10 * tenth + '0');
         n = tenth;
     } while (tenth != 0);
     *s = (char)(*s + min_offset);     
     *p = '\0';
}

-- 
pete

.