Conversion of base 10 to base 6

Question

I tried to convert a base 10 number to base 6 in C, but my code didn't pass 2 hidden test cases.

I can't find any logical error in it.

Could you?

//convert base 10 to base 6

#include<stdio.h>
int main()
{
   int num, rem = 0, i = 1, res = 0;
   scanf("%d", &num);
   while(num!=0)
   {
       rem = num%6;
       res = (rem*i)+res;
       num = num/6;
       i = i*10;
   }
   printf("%d",res);

}

Answer 1

Your solution will only work for a limited range of int.

Since base 6 will use more digits than a base 10 number, there will be a point where the base 10 number will generate a base 6 number that will not fit into an int, thus producing an overflow.

See this example.

One solution is to use strings to generate the base 6 number. The number is stored in a character array.

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main()
{
   const int maxdigits = 26;    /* max number of base 6 digits for a 64 bit int */
   int num=123456789, rem = 0;

   /* handle negative input */
   int tempnum = abs(num);

   int lastpos = maxdigits-1;

   /* initialize array, and initialize array to spaces */
   char res[maxdigits + 1];
   memset(res, ' ', maxdigits);

   /* null terminate */
   res[maxdigits] = 0;

   do 
   {
       rem = tempnum % 6;
       res[lastpos--] = rem + '0'; /* set this element to the character digit */
       tempnum /= 6;
   } while (tempnum > 0);
   printf("%s%s", (num < 0)?"-":"", res + lastpos + 1); /* print starting from the last digit added */
}

Output:

20130035113