How to add two numbers without using ++ or + or another arithmetic operator

Question

How do I add two numbers without using ++ or + or any other arithmetic operator?

It was a question asked a long time ago in some campus interview. Anyway, today some

Answer 1

If you're feeling comedic, there's always this spectacularly awful approach for adding two (relatively small) unsigned integers. No arithmetic operators anywhere in your code.

In C#:

static uint JokeAdder(uint a, uint b)
{
    string result = string.Format(string.Format("{{0,{0}}}{{1,{1}}}", a, b), null, null);
    return result.Length;
}

In C, using stdio (replace snprintf with _snprintf on Microsoft compilers):

#include <stdio.h>
unsigned int JokeAdder(unsigned int a, unsigned int b)
{
    return snprintf(NULL, 0, "%*.*s%*.*s", a, a, "", b, b, "");
}

Answer 2

For unsigned numbers, use the same addition algorithm as you learned in first class, but for base 2 instead of base 10. Example for 3+2 (base 10), i.e 11+10 in base 2:

   1         ‹--- carry bit
   0 1 1     ‹--- first operand (3)
 + 0 1 0     ‹--- second operand (2)
 -------
   1 0 1     ‹--- total sum (calculated in three steps)

Answer 3

int add_without_arithmatic(int a, int b)
{
    int sum;
    char *p;
    p = (char *)a;
    sum = (int)&p[b];
    printf("\nSum : %d",sum);
}

Answer 4

short int ripple_adder(short int a, short int b)
{
    short int i, c, s, ai, bi;

    c = s = 0;

    for (i=0; i<16; i++)
    {
        ai = a & 1;
        bi = b & 1;

        s |= (((ai ^ bi)^c) << i);
        c = (ai & bi) | (c & (ai ^ bi));

        a >>= 1;
        b >>= 1;
    }
    s |= (c << i);
    return s;
}

Answer 5

Well, to implement an equivalent with boolean operators is quite simple: you do a bit-by-bit sum (which is an XOR), with carry (which is an AND). Like this:

int sum(int value1, int value2)
{
    int result = 0;
    int carry = 0;
    for (int mask = 1; mask != 0; mask <<= 1)
    {
        int bit1 = value1 & mask;
        int bit2 = value2 & mask;
        result |= mask & (carry ^ bit1 ^ bit2);
        carry = ((bit1 & bit2) | (bit1 & carry) | (bit2 & carry)) << 1;
    }
    return result;
}

Answer 6

This can be done recursively:

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

    int sum = a ^ b; // add without carrying
    int carry = (a & b) << 1; // carry, but don’t add
    return add_without_arithm_recursively(sum, carry); // recurse
}

or iteratively:

int add_without_arithm_iteratively(int a, int b)
{
    int sum, carry;

    do 
    {
        sum = a ^ b; // add without carrying
        carry = (a & b) << 1; // carry, but don’t add

        a = sum;
        b = carry;
    } while (b != 0);

    return a;
}