#include <stdio.h>
      int add(int a, int b)
      {
	    if (!a)
		return b;
	    else
		return add((a & b) << 1, a ^ b);
      }

      int main()
      {
	    unsigned int a,b;
	    printf("Enter the two numbers: \n");

	    scanf("%d",&a);
	    scanf("%d",&b);
	    printf("Sum is: %d",add(a,b));
      }

Please explain me the code

Hi,
This is the code for a FULL ADDER circuit.

why cant we just or the two numbers

Hi Nitish,

If we will do the LOGICAL OR function then we will get
either 1 or 0. 
If we will do the BITWISE OR then we will get the largest of
the two.. 

For
 ex:
Let us take, First number as 2 and Second number as 3..
Then as per the first case we will get 1 as the output. 
10(2) || 11(3) -> 1(1)
As per the second case the output will be 3..
10(2) | 11(3) -> 11(3).
Ok

int sum(int num1,int num2)
{	
     for(int i=0;i<num2;i++)	
        num1++;	
     return num1;
}

even this gives the same ans as the above program gives...
just every one plz check it and tell me
#include<stdio.h>
#include<math.h>
int main()
{
int a,b,c;
printf("enter the nos");
scanf("%d %d",&b,&c);
a= (b^c);
printf("%d",a);
return 0;
}

#include<stdio.h>
int a,b,c;
{
printf("enter the two values");
scanf("%d","%d",&a,&b);
c=a||b;
}

prashant answer is wrong suppose add two similar numbers 
prashant answers will fail because addtion of two similar 
bits  according to the bitwise xor fails.

# include <stdio.h>
main()
{
int a=8,b=2;
a|=b;
printf("sum="+a);
}

#include<stdio.h>
main()
{
int a=5,b=4,c;
c=a||b;
printf("sum="+c);
}