Multiplication of two integers using bitwise operators

Question

How can I multipy two integers using bitwise operators?

I found an implementation here. Is there a better way of implementing multiplication?

For example: 2

Answer 1

In C# here is the implementation of the function.

    public static int Mul(int a, int b)
    {
        int r = 0;
        while (b != 0)
        {
            var temp = b & 1;

            if (temp!= 0)
            {
                r = r + a;
            }
            a= a << 1;
            b= b >> 1;
            if (temp == 0)
            {
                r = a;
                break;
            }
        }

        return r;
    }

Answer 2

    #include<stdio.h>
    void main()
    {
        int n, m, i, j, x, large, small, t1, m2, result, a1, a2, a3, c, c1, c2, r, r1, la, re;

        printf("Enter two numbers\n");
        scanf("%d%d", &n, &m);
        result = 0;

        if (n > m)
        {
            large = n;
            small = m;
        }
        else
        {
            large = m;
            small = n;
        }

        c = 0;

        while (small)
        {
            t1 = small;
            t1 &= 1;

            if (t1 == 1)
            {
                printf("\n %d", large);
                la = large;
                re = result;
                m2 = 0;
                r1 = 1;
                while (re || la || c)
                {
                    a2 = la;
                    a2 &= 1;
                    a3 = re;
                    a3 &= 1;

                    c1 = a2 & a3;
                    r = a3 ^ a2;

                    c2 =r & c;
                    r ^= c;
                    if (c1 || c2)
                        c = 1;
                    else
                        c = 0;

                    result &= ~r1;
                    x = r;
                    m2 >>= 1;
                    while (m2)
                    {
                        r <<= 1;
                        m2 >>= 1;
                    }

                    result |= r;
                    la >>= 1;
                    re >>= 1;
                    r1 <<= 1;
                    m2 = r1;
                }

            }
            large <<= 1;
            small >>= 1;

        }
        printf("\n%dX%d= %d\n", n, m, result);
    }

Answer 3

This one is purely with bit-wise operations.

public int bitwiseMultiply(int a, int b) {
    if (a ==0  || b == 0) {
        return 0;
    }

    if (a == 1) {
        return b;
    }
    else
        if (b == 1) {
            return a;
        }


    int result = 0; // Not needed, just for test
    int initA = a;
    boolean isORNeeded = false;
    while (b != 0 ) {

        if (b == 1) {
            break;
        }

        if ((b & 1) == 1) { // Carry needed, odd number
            result += initA; // Test, not needed
            isORNeeded = true;
        }

        a <<= 1; // Double the a
        b >>= 1; // Half the b
        System.out.println("a=["+a+"], b=["+b+"], result=["+result+"]");
    }

    return (isORNeeded ? (a | initA) : a); // a + result;
}

Answer 4

Assembly algorithm: This follows directly from the fact that ax*7 = (ax*8)-ax.

mov     bx, ax          ;Save AX*1
shl     ax, 1           ;AX := AX*2
shl     ax, 1           ;AX := AX*4
shl     ax, 1           ;AX := AX*8
sub     ax, bx          ;AX := AX*7

Every shift step is a multiplication by 2

Answer 5

Below is one possible solution for multiplication of two integers using bitwise operators.

#include <stdio.h>
#define INT_BITS 32

int multiply(int a, int b)
{
    int pos1, pos2, res;
    for (pos1 = 0; pos1 < INT_BITS; pos1++)
      {
        for (pos2 = 0; pos2 < INT_BITS; pos2++)
          {
            /* Find the bits set in both numbers and add their
             * corresponding powers of 2. 
             */
            if ((a & (1 << pos1)) && (b & (1 << pos2)))
              {
                res = res + (1 << (pos1 + pos2));
                // res = res + ((1 << pos1) << pos2);
                /* Both the above statements calculating res are same */
              }
          }
      }

  return res;
}


int main()
{
    int num1, num2, product;
    printf("Enter two numbers to be multiplied:");
    scanf("%d %d", &num1, &num2);
    product = multiply(num1, num2);
    printf("Product of %d and %d: %d\n", num1, num2, product);
    return 0;
}

Function multiply() can be slighted changed as below using Shiv's Add() function:

int Add(int x, int y)
{
    if (y == 0)
        return x;
    else
        return Add( x ^ y, (x & y) << 1);
}

int multiply(int a, int b)
{
    int pos1, pos2, res, temp;
    for (pos1 = 0; pos1 < INT_BITS; pos1++)
      {
        for (pos2 = 0; pos2 < INT_BITS; pos2++)
          {
            /* Find the bits set in both numbers and add their
             * corresponding powers of 2. 
             */
            if ((a & (1 << pos1)) && (b & (1 << pos2)))
              {
                temp = (1 << pos1) << pos2;
                res = Add(res, temp);
              }
          }
      }

  return res;
}

Answer 6

#include <stdio.h>

int main(void)
{
    int a, b, result;
    printf("Enter the numbers to be multiplied:");
    scanf("%d%d", &a, &b);       // a > b
    result = 0;
    while (b != 0)               // Iterate the loop till b == 0
    {
        if (b & 1)               // Bitwise & of the value of b with 1
        {
            result = result + a;  // Add a to result if b is odd .
        }
        a <<= 1;                    // Left shifting the value contained in 'a' by 1
                                  // Multiplies a by 2 for each loop
        b >>= 1;                    // Right shifting the value contained in 'b' by 1.
    }

    printf("Result: %d\n",result);
}

Source