[vox-tech] [OT] Binary Representation Challenge

[Tim Riley]

On Tue, 2005-09-20 at 12:02, Alex Mandel wrote:
> I realize this might be not be a challenge for some of you.
> --
> I need to make a list of all possible permutations given 9 options and 
> that you can choose any number of options at once.
> I've figured out using nCr statistics that this is 511 choices, but now 
> I need to represent them in 2^9 binary code: 000000001, 000000010 etc

The following code should work:

/* binary_count.c					*/
/* ---------------------------------------------------- */
/* By Tim Riley						*/
/* ---------------------------------------------------- */
/* Compile: cc -Wall binary_count.c -o binary_count	*/
/* ---------------------------------------------------- */

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

#define NUMBER_BINARY_DIGITS	16

/* From "The C Programming Language" by Kernighan and Ritchie */
/* ---------------------------------------------------------- */
unsigned getbits( unsigned integer, unsigned position, unsigned n )
{
	return ( integer >> ( position + 1 - n ) ) & ~( ~0 << n );
}

char *integer2binary( unsigned integer )
{
	static char binary[ NUMBER_BINARY_DIGITS + 1 ] = {0};
	char *pointer;
	int i;

	pointer = binary;
	for ( i = 0; i < NUMBER_BINARY_DIGITS; i++, pointer++ )
		*pointer = '0';

	pointer = binary + NUMBER_BINARY_DIGITS - 1;

	for( i = 0; i < NUMBER_BINARY_DIGITS; i++, pointer-- )
	{
		if ( getbits( integer, i, 1 ) ) *pointer = '1';
		
	}

	return binary;
}

int main( int argc, char **argv )
{
	unsigned count_up_to;
	unsigned count;

	if ( argc != 2 )
	{
		fprintf( stderr, "Usage: %s count_up_to\n", argv[ 0 ] );
		return 1;
	}

	count_up_to = atoi( argv[ 1 ] );

	for ( count = 0; count < count_up_to; count++ )
		printf( "%s\n", integer2binary( count ) );

	return 0;
}


<snip>