Adding a two number without using + sign

Question

I am 9 grade, My math teacher asked me to add numbers with out using + sign in C program.

I tried a - (-b) = a + b; but my math teacher want some other option.

He probably means bitwise then? Off the top of my head: AND to spot bits that'll carry, XOR to add; then shift the AND result left and repeat adding that to the XOR value until there's no AND left? — Rup, Sep 25 '13 at 17:00
Perhaps try something more... iterative. ( think a `while` ) — , Sep 25 '13 at 17:00
How about [Addition of two integers using bitwise operators](http://stackoverflow.com/q/4068033/1168156) ? — LihO, Sep 25 '13 at 17:00
Wow, a nice question. I'm gonna learn much with this question — Sri Harsha Chilakapati, Sep 25 '13 at 17:07
You should really ask your prof to be more specific on the constraints. The logarithm-based approach below is elegant, but seems a bit too "easy". I'm suspecting your prof, given that this is a programming course, wants you to use a boolean-logic based approach, rather than an algebraic approach. +1 to all the solutions below! — Cloud, Sep 25 '13 at 17:26
Although the question is rather interesting, this is asking to do his homework. — Rui Marques, Sep 25 '13 at 18:37

score 23 · Answer 1 · edited Sep 25 '13 at 17:15

Use this function in your c program

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

        // Sum of bits of "a" and "b" where at least one of the bits is not set
        a = a ^ b;

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

zavg · Answer 2 · 2013-09-25T18:59:07.573

12

Use bitwise ^ and & operators and recursion

int add(int x, int y){
    return y == 0 ? x : add( x ^ y, (x & y) << 1);
}

P.S.: It is the recursive version of an algorithm proposed by vikas.

edited Sep 25 '13 at 18:59

answered Sep 25 '13 at 17:04

zavg

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score 8 · Answer 3 · answered Sep 25 '13 at 17:05

In Java using recursion-

public static int add(int a, int b){

    if(b == 0) return a;
    int sum = a ^ b;
    int carry = (a & b) << 1;

    return add(sum, carry);

}

In C-

int add(int a, int b){

    if(b == 0) return a;
    int sum = a ^ b;
    int carry = (a & b) << 1;

    return add(sum, carry);
}

score 7 · Accepted Answer · edited Sep 25 '13 at 17:29

7

Using Anti Log() you can do that

Temp= Anti Log(a)* Anti Log(b);

a+b value is equals to log(Temp);

Works for integers not for double.

edited Sep 25 '13 at 17:29

Gangadhar

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answered Sep 25 '13 at 17:00

cvsr.sarma

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What is the difference? – Sep 25 '13 at 17:03
I like this logic. It worked for integers. – Sep 25 '13 at 17:06
Difference its not exact equal. there is some difference after decimal point – cvsr.sarma Sep 25 '13 at 17:13
1

for a-b = log(antilog(a)/antlog(b)) for a*b=log(antiLog(a)+antLog(b)) for a/b= log((antilog(a)-(antilog(b))) – cvsr.sarma Sep 25 '13 at 17:21
@sarmachinta...I don't think your teacher expecting this solution to use Log. Better to use the one mentioned by vikas. – surender8388 Sep 25 '13 at 17:36
6

"Anti Log" is a pretty strange name, considering that logarithm _is itself defined as_ the inverse of exponentiation... – leftaroundabout Sep 25 '13 at 18:37

score 2 · Answer 5 · answered Sep 25 '13 at 19:09

#include<stdio.h>

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

int main() {
    printf("%d", add(5, 17));
    getchar();
}

Not portable but doesn't use the "+" character. This casts a to a char pointer, adds b to it with [] and then casts it back to an int.

Adding a two number without using + sign

5 Answers5