Decimal to binary algorithm in C - 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);

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

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 5

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 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;
}