Decimal to binary algorithm in C

Question

I am trying to use the following algorithm to convert a decimal number to a binary number in C. I don't understand why it doesn't work properly for some inputs (e.g. for 1993 I get 1420076519).

int aux=x;
long bin=0;
while (aux>0)
{
    bin=bin*10+aux%2;
    aux=aux/2;
}
printf("%d in decimal is %ld in binary.", x, bin);

Solution 1

When you print a long you dont print the binary. The best way to convert to binary or show the binary representation of a decimal number is by storing it in a string. Bellow is a solution offered in a another SO answer

void getBin(int num, char *str)
{
  *(str+5) = '\0';
  int mask = 0x10 << 1;
  while(mask >>= 1)
    *str++ = !!(mask & num) + '0';
}

Solution 2

If you know the algorithm there's no reason not to use itoa

http://www.cplusplus.com/reference/clibrary/cstdlib/itoa/

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

int main ()
{
  int n;
  char output[100];

  printf("Enter a number: ");
  scanf("%d", &n);

  itoa(n, output, 2); //2 means base two, you can put any other number here

  printf("The number %d is %s in binary.", n, output);

  return 0;
}

Solution 3

How does the conversion works?

/* Example: 
   125(10) -----> ?(2)                     125  |_2
                                            -1-   62  |_2
                                                  -0-   31 |_2
                                                        -1-  15 |_2
                                                             -1-  7 |_2
                                                                 -1-  3 |_2
                                                                     -1-  1 */

So in this example the binary number for 125(10) is 1111101(2), and this is the process I describe in my function.

/* Functions declaration (Prototype) */

 int wordCalculator( int * const word, long int number, int base );
    int main( void )
        {
            int i, base;
            int word[ 32 ];
            unsigned long int number;

            printf( "Enter the decimal number to be converted: " );
            scanf( "%ld", &number );
            printf( "\nEnter the new base: " );
            scanf( "%d", &base );

            i = wordCalculator( word, number, base );

            printf( "The number is: " );

            for(; i >= 0; i--){

                if ( word[ i ] <= 9)
                    printf( "%d", word[ i ] );

                else
                    /* 65 represents A in ASCII code. */
                    printf( "%c", ( 65 - 10 + word[ i ] ) );
            }

            printf( "\n" );
        }

        int wordCalculator( int * const word, long int number, int base )
        {
            unsigned long int result = number;
            int i, difference;

            i = 0;
            do{
                difference = result % base;
                result /= base;
                *( word + i ) = difference;
                i++;

                if ( result < base )
                    *( word + i ) = result;

            } while( result >= base );

            return i;

        }