Add two integers using only bitwise operators?

In C#, is it possible to perform a sum of two 32-bit integers without using things like if..else, loops etc?

That is, can it be done using only the bitwise operations OR (|), AND (&), XOR (^), NOT (!), shift left (<<) and shift right (>>)?

Here is an example for your amusement

unsigned int myAdd(unsigned int a, unsigned int b)
{
    unsigned int carry = a & b;
    unsigned int result = a ^ b;
    while(carry != 0)
    {
        unsigned int shiftedcarry = carry << 1;
        carry = result & shiftedcarry;
        result ^= shiftedcarry;
    }
    return result;
}

The loop could be unrolled. Number of times it executes, depends on the number of bits set in operands, but it's never greater than the width of unsigned int. Once carry becomes 0, next iterations don't change anything.

Try this:

    private int add(int a, int b) {
        if(b == 0)
            return a;

        return add( a ^ b, (a & b) << 1);
    }

Edit: Corrected if statement

Think about how addition happens bit by bit. Shift the values to get each bit of each operand in turn, then look at the four possible values for the two bits and work out what the result bit should be and whether there's a carry bit to worry about. Then see how the result and carry can be caculated using the bitwise ops.

static int binaryadd(int x, int y)
{
  while (x != 0)
  {
    int c = y & x;
    y = y ^ x; 
    x = c << 1;             
  }
  return y;
}

public static int getSum(int p, int q)
{
    int carry=0, result =0;
    for(int i=0; i<32; i++)
    {
        int n1 = (p & (1<<(i)))>>(i); //find the nth bit of p
        int n2 = (q & (1<<(i)))>>(i); //find the nth bit of q
        
        int s = n1 ^ n2 ^ carry; //sum of bits
        carry = (carry==0) ? (n1&n2): (n1 | n2); //calculate the carry for next step
        result = result | (s<<(i)); //calculate resultant bit
    }
    
    return result;
}

Taking 32 bit as int takes 32 bit. Thanks!!!

int Add(int a, int b)
{
      int result = 0,
          // carry now contains common set bits of "a" and "b"
          carry = a & b;

      if (Convert.ToBoolean(carry))
      {
          // Sum of bits of "a" and "b" where at least one 
          // of the bits is not set
          result = a ^ b;

          // carry is shifted by one so that adding it 
          // to "a" gives the required sum
          carry = carry << 1;

          result = add(carry, result);
      }
      else
      {
          result = a ^ b;
      }

      return result;
}

Sum of two bits can be performed using the XOR ^ operator and carry bit can be obtained by using AND & operator. Provided a and b don't have set bits at the same position, then using ^ operator gives the sum of a and b.

Comments from geeksforgeeks

        int  b =  25;
        for (int t = 128; t > 0; t = t / 2)
        {
            if ((b & t) != 0) Console.Write("1 ");
            if ((b & t) == 0) Console.Write("0 ");
        }
        Console.WriteLine();
        //b = (sbyte)~b;
        int e = 22;
        for (int t = 128; t > 0; t = t / 2)
        {
            if ((e & t) != 0) Console.Write("1 ");
            if ((e & t) == 0) Console.Write("0 ");
        }
        Console.WriteLine();
        int c = b | e;
        for (int t = 128; t > 0; t = t / 2)
        {
            if ((c & t) != 0) Console.Write("1 ");
            if ((c & t) == 0) Console.Write("0 ");
        }