Thread: decimal to binary

Hi Guys
Ive searched the forum but havent found much on decimal to binary so i came up with my own algorithm. However im encountering a seg fault and dont know where the problem lies.
My algorithm to convert decimal to binary is:
get the number
find the highest power of 2
compare the number with highest power
if number greater the highest power print a 1
adjust the new number
adjust the new power
end when there is a zero.


Cheers

Code:

void dec2bin( void ) {
{
	int num;
	int highestpower=1;
	int i;
	printf( "Enter a decimal number: " );
	scanf( "%d", &num );
		for ( i=1; i<=num; i++ ) {
		if ( highestpower <= num )
			highestpower = intpow( 2, i );
		else if ( highestpower > num )
			break;
	}
		while ( num != 0 ) {
		if ( num >= highestpower ) {
			putchar( '1' );
			num = num - highestpower;
			highestpower = highestpower - intpow( 2, sqrt(highestpower)-1);
		}
		else  {	
			putchar( '0' );
			highestpower = highestpower - intpow( 2, sqrt(highestpower)-1);
		}
	}
}

int intpow( int base, int power )
{
	if ( power == 0 )
		return 1;
	else
		return base * intpow( base, power-1 );
}

Code:

#include <stdio.h>

void dec2bin( int encounteredOne ) 
{
  int theNum;
  int curBit;;
  
  printf( "Enter a number: " );
  fflush( stdout );
  scanf( "%d", &theNum );
  
  for( curBit = ( sizeof( theNum ) * 8 ) - 1; curBit >= 0; curBit-- ) {
    
    if( encounteredOne == 0 ) 
      if( theNum>>curBit & 0x01 == 0x01 ) 
	encounteredOne = 1;
    if( encounteredOne == 1 ) {
      printf( "%d", theNum>>curBit & 0x01 );
      if( curBit % 8 == 0 ) printf( " " );
    }
    
    
  }
  
  printf( "\n" );
  
}


int main( ) 
{
  dec2bin( 1 ); // use 0 if you want it to truncate it (i.e. 000101 becomes 101)
  return 0;
}

Slight correction... to get no. bits required to store target you need:

Code:

j = log10(target) / log10(2) + 1;

and you are trying to be a bit to clever with your for-loop. This version gives the right answer:

Code:

  for(i=(j-1);i >= 0; i--)
  /* or 'for(i=j;i--; )' would also work but is a bit too cryptic for me to get my head round */
  {
    if(target & (1 << i))
      putc('1', stdout);
    else
      putc('0', stdout);
  }

thanks guys for the replies.

but just unsure how a number gets converted to binary
for example the decimal number is 20 which is 010100 in binary. How does the bitwise recognise the decimal number 20 and convert it?

the code you have written shifts from the left so does that start with the 0 in the decimal number 20?

if so then its 0 & 0 = 0
then it goes to 2, 2 & 0 = ??

im unsure of what is taking place
cheers

hmm so the moment i enter a decimal number its converted to binary already?

so then just use bitwise operaters and compare and print out the values?

is that the logic here?

Yep, bitwise is the way to go...and here you go for 32-bit binary!!!

Code:

#include <stdio.h>
/*------------------------------------------------------------*/
/* Decimal to binary conversion */

void Decimal_2_Binary(unsigned int x, unsigned int c[32])
{
	int bit;		

	for(bit = 0; bit < 32; bit++)
	{
		c[bit] = ((x >> bit) & 1);
	}
}


/*------------------------------------------------------------*/

int main()
{
    int bit;
    unsigned int x;
    unsigned int c[32];		

	
    printf("\nEnter an integer number smaller than 4,294,967,295: ");
    scanf("%u", &x);
    printf("\nConverting decimal to binary:\n");

    Decimal_2_Binary(x, c);

    printf("	Binary == ");

    for(bit = 31; bit >= 0; bit--)
    {
	if(bit == 27 || bit == 23 || bit == 19 || bit == 15 || bit == 11 || bit == 7 || bit == 3)
	{
		printf(" ");
	}


	printf("%u", c[bit]);
    }

	printf("\n\n\n");
	return 0;
}