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

I came here looking for this question and I find Zengr's answer correct. Thanks Zengr! But there is one modification I would want to see which is getting rid of the '+' operator in his code. This should make multiplication of two arbitrary numbers using NO ARITHMETIC OPERATORS but all bitwise.

Zengr's solution first:

#include<stdio.h>
main()
{
   int a,b,result;     
   printf("nEnter 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&01)                // Bitwise &  of the value of b with 01
        {
          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("nResult:%d",result);
}

My Answer would be:

#include<stdio.h>
main()
{
   int a,b,result;     
   printf("nEnter 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&01)                // Bitwise &  of the value of b with 01
        {
          result=add(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("nResult:%d",result);
}

where I would write add() as:

int Add(int x, int y)
{
    // Iterate till there is no carry  
    while (y != 0)
    {
        // carry now contains common set bits of x and y
        int carry = x & y;  

        // Sum of bits of x and y where at least one of the bits is not set
        x = x ^ y; 

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

or recursively adding as:

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

source for addition code: http://www.geeksforgeeks.org/add-two-numbers-without-using-arithmetic-operators/