Subtracting two numbers without using '-' operator

Question

hi, i tried with the following code , but i can't understand why it's giving me wrong answer. i am computing the 2's complement and adding with another no.

#include <stdio.h>

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

int sub(int a, int b) // add a with b's 2's complement.
{
    return (add(a, add(~b, 1)));
}

int main() {
    int a, b, res;
    a = 3, b = 1;
    res = sub(a, b);
    printf("%d\n", res);
    return 0;
}

Answer 1

Here is better solution
http://stackoverflow.com/questions/1149929/

#include <stdlib.h> /* atoi() */
#include <stdio.h>  /* (f)printf */
#include <assert.h> /* assert() */

int add(int x, int y) {
    int carry = 0;
    int result = 0;
    int i;

    for(i = 0; i < 32; ++i) {
        int a = (x >> i) & 1;
        int b = (y >> i) & 1;
        result |= ((a ^ b) ^ carry) << i;
        carry = (a & b) | (b & carry) | (carry & a);
    }

    return result;
}

int negate(int x) {
    return add(~x, 1);
}

int subtract(int x, int y) {
    return add(x, negate(y));
}

int is_even(int n) {
    return !(n & 1);
}

int divide_by_two(int n) {
    return n >> 1;
}

int multiply_by_two(int n) {
    return n << 1;
}

int multiply(int x, int y) {
    int result = 0;

    if(x < 0 && y < 0) {
        return multiply(negate(x), negate(y));
    }

    if(x >= 0 && y < 0) {
        return multiply(y, x);
    }

    while(y > 0) {
        if(is_even(y)) {
                x = multiply_by_two(x);
                y = divide_by_two(y);
        } else {
                result = add(result, x);
                y = add(y, -1);
        }
    }

    return result;
}

int main(int argc, char **argv) {
    int from = -100, to = 100;
    int i, j;

    for(i = from; i <= to; ++i) {
        assert(0 - i == negate(i));
        assert(((i % 2) == 0) == is_even(i));
        assert(i * 2 == multiply_by_two(i));
        if(is_even(i)) {
                assert(i / 2 == divide_by_two(i));
        }
    }

    for(i = from; i <= to; ++i) {
        for(j = from; j <= to; ++j) {
                assert(i + j == add(i, j));
                assert(i - j == subtract(i, j));
                assert(i * j == multiply(i, j));
        }
    }

    return 0;
}

Answer 2

i used a different add() function as suggested by NullUserException, it works now:

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

Answer 3

Considering how negative numbers are represented, the following will compute a - b:

int a, b, c;
// assign to a and b
c = a + (~b + 1); // () not needed, just to show the point

as the OP already noted:) This moves the attention to your add implementation, that is of course wrong. The following is an odd way to do it (just since other better ways are already given)

int add1(int a, int b, int *c)
{
  int r = *c & 1;
  a &= 1; b &= 1;
  *c = a&b | a&r | b&r;
  return a^b^r;
}
int inv(int a)
{
  int i, r = 0;
  for(i = 0; i < sizeof(int)*8; i++)
  {
    r = r<<1 | (a&1);
    a >>= 1;
  }
  return r<<1;
}
int add(int a, int b)
{
  int r = 0, i;
  int c = 0;
  for(i=0; i < sizeof(int)*8; i++)
  {
    r |= add1(a>>i, b>>i, &c);
    r <<= 1;
  }
  return inv(r);
}

int sub(int a, int b)
{
  return add(a, add(~b, 1));
}

(keeping the same idea the code can be made better, just too tired to do it finer)