What is the best way to add two numbers without using the + operator?

Question

A friend and I are going back and forth with brain-teasers and I have no idea how to solve this one. My assumption is that it's possible with some bitwise operators, but not sure.

Answer 1

Why not just incremet the first number as often, as the second number?

Answer 2

Cheat. You could negate the number and subract it from the first :)

Failing that, look up how a binary adder works. :)

EDIT: Ah, saw your comment after I posted.

Details of binary addition are here.

Answer 3

In C, with bitwise operators:

#include<stdio.h>

int add(int x, int y) {
    int a, b;
    do {
        a = x & y;
        b = x ^ y;
        x = a << 1;
        y = b;
    } while (a);
    return b;
}


int main( void ){
    printf( "2 + 3 = %d", add(2,3));
    return 0;
}

Answer 4

The reason ADD is implememted in assembler as a single instruction, rather than as some combination of bitwise operations, is that it is hard to do. You have to worry about the carries from a given low order bit to the next higher order bit. This is stuff that the machines do in hardware fast, but that even with C, you can't do in software fast.

Answer 5

Define "best". Here's a python version:

len(range(x)+range(y))

The + there is list catenation, not addition.

Answer 6

Adding two integers is not that difficult; there are many examples of binary addition online.

A more challenging problem is floating point numbers! There's an example at http://pages.cs.wisc.edu/~smoler/x86text/lect.notes/arith.flpt.html

Answer 7

Note, this would be for an adder known as a ripple-carry adder, which works, but does not perform optimally. Most binary adders built into hardware are a form of fast adder such as a carry-look-ahead adder.

My ripple-carry adder works for both unsigned and 2's complement integers if you set carry_in to 0, and 1's complement integers if carry_in is set to 1. I also added flags to show underflow or overflow on the addition.

#define BIT_LEN 32
#define ADD_OK 0
#define ADD_UNDERFLOW 1
#define ADD_OVERFLOW 2

int ripple_add(int a, int b, char carry_in, char* flags) {
    int result = 0;
    int current_bit_position = 0;
    char a_bit = 0, b_bit = 0, result_bit = 0;

    while ((a || b) && current_bit_position < BIT_LEN) {
        a_bit = a & 1;
        b_bit = b & 1;
        result_bit = (a_bit ^ b_bit ^ carry_in);
        result |= result_bit << current_bit_position++;
        carry_in = (a_bit & b_bit) | (a_bit & carry_in) | (b_bit & carry_in);
        a >>= 1;
        b >>= 1;
    }

    if (current_bit_position < BIT_LEN) {
        *flags = ADD_OK;
    }
    else if (a_bit & b_bit & ~result_bit) {
        *flags = ADD_UNDERFLOW;
    }
    else if (~a_bit & ~b_bit & result_bit) {
        *flags = ADD_OVERFLOW;
    }
    else {
        *flags = ADD_OK;
    }

    return result;
}

Answer 8

No + right?

int add(int a, int b) 
{
   return -(-a) - (-b);
}

Answer 9

an abacus will do this quite well, and it doesn't use any electricity!

Answer 10

int add(int a, int b) { const char *c=0; return &(&c[a])[b]; }

Answer 11

CMS Can you please explain the logic? I am unable to understand your logic

Answer 12

CMS's add() function is beautiful. It should not be sullied by unary negation (a non-bitwise operation, tantamount to using addition: -y==(~y)+1). So here's a subtraction function using the same bitwise-only design:

int sub(int x, int y) {
    unsigned a, b;
    do {
        a = ~x & y;
        b =  x ^ y;
        x = b;
        y = a << 1;
    } while (a);
    return b;
}