Thread: Shift all '1' bits in a number to the left end

Hello,

I'm working on Practical C Programming exercise 11-6 and have a solution. I just feel like it's kind of messy, so I wanted to see if others had feedback on a cleaner, easier way. Mind you, we haven't reach pointers yet in the book. Thanks!

Code:

/*************************************************************
	pcp11_6	--	Takes the 1 bits in a number and shifts them to the left
				end.

*************************************************************/
#include <stdio.h>

char			line[100];		/* user input */
unsigned char 	num;			/* number from user input (8-bit)*/
int			ones_count;	/* number of 1s to shift */
unsigned char	new_num;		/* new number with shifted bits */
int			i,j;		        /* counters */
int			k;		        /* bit shift counter */

int main(void)
{
	//initialize vars
	ones_count = 0;
	k = 7;

	// Get user input and store in integer
	printf("Please enter a number (1 to 255): ");
	fgets(line, sizeof(line), stdin);
	sscanf(line, "%d", &num);

	// Count number of 1s in the 8-bit number
	for(i = 0x80; i > 0; i = (i >> 1))
	{
		if(num & i)
		{
			++ones_count;
		}
	}

     // Create new number, bit-by-bit
	for(i = 0x80; i > 0; i = (i >> 1))
	{
               // Push all of the 1s to the left side of the new number
		while(ones_count != 0)
		{
			new_num |= 1 << k;
			i = i >> 1;
			--ones_count;
			--k;
		}
               // Fill in the rest of the number with 0s
		new_num |= 0 << k;
		--k;
	}
	printf("\nNew number is = %d.\n", new_num);

	return 0;
}