Sum binary numbers

[lobodelbosque]

Hello:

I wrote the following code to sum two binary numbers of the same length:

public class BinarySum {


	public static void main(String[] args) {
		
		int[] a = {0,1,1,1,1,1};
		int[] b = {0,1,1,0,1,1};
		int[] c = new int[a.length + 1];
		int n = a.length + 1;
		int m = a.length;
		int aux = 0;
		int i;
		
		for(i=1 ; i <= m; i++)
		{
			if(a[m - i] + b[m - i] == 2)
			{		
				c[n-i]= 0 + aux;
				aux = 1;
				continue;
				
			}
			
			else
			{
				if(a[m - i] + b[m - i] + aux == 2)
				{
					c[n-i] = 0;
					aux=1;
					continue;
				}						
				
			
				else
				{
					c[n-i] = a[a.length - i] + b[b.length - i] + aux;
					aux=0;
				}
				
				
			}
			
			
		}
		if(aux == 1)
			c[0]=1;
		else
			c[0]=0;
		for(i = 0 ; i < c.length ; i++)
		System.out.print(c[i]);
		
		
		
		
		

	}

}

I feel it cant be shorter, i mean a way more intelligent to accomplish the task.
Anyone have another idea?

Thanks.

[pbrockway2]

for ndx = end downto 0
    s = a[ndx] + b[ndx] + carry
    sum[ndx + 1] = s % 2
    carry = s / 2
sum[0] = carry

or something...

[lobodelbosque]

I see thats a very shorter way.
Thanks.

[pbrockway2]

You're welcome. It's just a pseudocode description of the ordinary way we do addition (with 10 for 2): "mod" and "div" are actually old friends although they're not usually given names when teaching children to add.