Add two unsigned integers without using '+'

I found this very interesting discussion (and the program) to add two numbers without the use of the arithmetic operator '+'. The obvious guess would be to use the Bitwise operators.

The logic behind the addition operation using the bitwise operators is as follows:

integer1 =  3  = 0011b
integer2 =  5  = 0101b

    first operation        second operation      third operation
        0011                0011                 shift by one
        0101                0101                   
       ______              ______
 XOR    0110           AND  0001  ------>        <<1   0010

    Again...  
                   first operation   second operation   third operation
previous XOR        0110                0110            shift by one
previous <<1        0010                0010
                   ______              ______
               XOR  0100            AND 0010  ------>  <<1   0100

    Again...  
                   first operation  second operation    third operation
previous XOR        0100                0100            shift by one
previous <<1        0100                0100
                   ______              ______
                XOR 0000           AND  0100  ------>  <<1   1000

    Again...  
                   first operation   second operation   
previous XOR        0000               0000            
previous <<1        1000               1000
                    ______            ______
                XOR 1000          AND  0000  
When AND iguals 0 the result of XOR is iqual to the sum of the two integers

The program is as follows:

//Program to add two numbers by using boolean operators
//Ref: http://www.daniweb.com/forums/thread84950.html
//Program tested on Microsoft Visual Studio 2008 - Zahid Ghadialy
#include<iostream>

using namespace std;

unsigned long add(unsigned integer1, unsigned integer2)
{
  unsigned long xor, and, temp;

  and = integer1 & integer2; /* Obtain the carry bits */
  xor = integer1 ^ integer2; /* resulting bits */
    
  while(and != 0 ) /* stop when carry bits are gone */
  {
    and <<= 1; /* shifting the carry bits one space */
    temp = xor ^ and; /* hold the new xor result bits*/
    and &= xor; /* clear the previous carry bits and assign the new carry bits */
    xor = temp; /* resulting bits */
  }
  return xor; /* final result */
}

int main()
{
  int num1 = 13, num2 = 27;

  cout << num1 << " + " << num2 << " = " <<add(num1, num2) << endl;

  return 0;
}

The output is:

13 + 27 = 40

See the original discussion here.