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 (eg 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);

Answer 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';
}

Answer 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;
}

Answer 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;

        }

Answer 4

You should be using strings to store binary number. Following code should work for you.

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

char *decimal_to_binary(int);

main()
{
   int n, c, k;
   char *pointer;

   printf("Enter an integer in decimal number system\n");
   scanf("%d",&n);

   pointer = decimal_to_binary(n);
   printf("Binary string of %d is: %s\n", n, pointer);

   free(pointer);

   return 0;
}

char *decimal_to_binary(int n)
{
   int c, d, count;
   char *pointer;

   count = 0;
   pointer = (char*)malloc(32+1);

   if ( pointer == NULL )
      exit(EXIT_FAILURE);

   for ( c = 31 ; c >= 0 ; c-- )
   {
      d = n >> c;

      if ( d & 1 )
         *(pointer+count) = 1 + '0';
      else
         *(pointer+count) = 0 + '0';

      count++;
   }
   *(pointer+count) = '\0';

   return  pointer;
}

Answer 5

I think the shortest answer is

char* getBinary(int n,char *s)
{
  while(n>0)
  {
    *s=(n&1)+'0';
    s++;
    n>>=1;
  }
  *s='\0';
  return s;
}

In the called function print it in reverse way .. because storing is done LSB to MSB But we have to print MSB first then LSB

Answer 6

You can use the below algorithm to convert Decimal number to Binary number system .

#include <stdio.h>  

int main()  
{  
    long long decimal, tempDecimal, binary;  
    int rem, place = 1;  

    binary = 0;  

    /* 
     * Reads decimal number from user 
     */  
    printf("Enter any decimal number: ");  
    scanf("%lld", &decimal);  
    tempDecimal = decimal;  

    /* 
     * Converts the decimal number to binary number 
     */  
    while(tempDecimal!=0)  
    {  
        rem = tempDecimal % 2;  

        binary = (rem * place) + binary;  

        tempDecimal /= 2;  
        place *= 10;  
    }  

    printf("\nDecimal number = %lld\n", decimal);  
    printf("Binary number = %lld", binary);  

    return 0;  
}  

Answer 7

This is a recursive solution that i wrote, it is simple and works fine.

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

int  printBinary(int N)
{
    if(N < 0){errno = EINVAL; return -1;}

    if(N == 0)
        printf("0");
    else if(N == 1)
        printf("1");
    else
    {
        printBinary(N/2);
        printf("%d", N%2);
    }

    return 0;
}

int main(int argc, char* argv[])
{
    if(argc < 2)
    {
        fprintf(stderr, "usage: %s NUM\nWhere NUM is an integer number\n", argv[0]);
        exit(EXIT_FAILURE);
    }
    errno = 0;


    long NUM = strtol(argv[1], NULL, 10);
    if(NUM == 0 && errno != 0)
    {
        perror("Error during number acquisition: ");
        exit(EXIT_FAILURE);
    }

    if(NUM < 0)
    {
        printf("-");
        printBinary(-NUM);
    }
    else
        printBinary(NUM);

    printf("\n");

    return 0;
}