Compute modulus division by a power-of-2-number

Compute n modulo d without division(/) and modulo(%) operators, where d is a power of 2 number.

Let ith bit from right is set in d. For getting n modulus d, we just need to return 0 to i-1 (from right) bits of n as they are and other bits as 0.

For example if n = 6 (00..110) and d = 4(00..100). Last set bit in d is at position 3 (from right side). So we need to return last two bits of n as they are and other bits as 0, i.e., 00..010.

Now doing it is so easy, guess it….

Yes, you have guessing it right. See the below program.

C++

#include<stdio.h>

// This function will return n % d.
// d must be one of: 1, 2, 4, 8, 16, 32, … 
unsigned int getModulo(unsigned int n, 
                       unsigned int d)
{
return ( n & (d - 1) );
}         

// Driver Code
int main()
{
unsigned int n = 6;

// d must be a power of 2
unsigned int d = 4; 
printf("%u moduo %u is %u", n, d, getModulo(n, d));

getchar();
return 0;
}     

Java

// Java code for Compute modulus division by 
// a power-of-2-number
class GFG {
    
    // This function will return n % d.
    // d must be one of: 1, 2, 4, 8, 16, 32,
    static int getModulo(int n, int d)
    {
        return ( n & (d-1) );
    }     
    
    // Driver Code
    public static void main(String[] args)
    {
        int n = 6;
        
        /*d must be a power of 2*/
        int d = 4; 
        
        System.out.println(n+" moduo " + d + 
                    " is " + getModulo(n, d));
    }
} 

// This code is contributed 
// by Smitha Dinesh Semwal.

Python3

# Python code to demonstrate
# modulus division by power of 2


# This function will
# return n % d.
# d must be one of:
# 1, 2, 4, 8, 16, 32, … 
def getModulo(n, d):

    return ( n & (d-1) )
         
# Driver program to
# test above function 
n = 6

#d must be a power of 2
d = 4 
print(n,"moduo",d,"is",
      getModulo(n, d))

# This code is contributed by 
# Smitha Dinesh Semwal

C#

// C# code for Compute modulus
// division by a power-of-2-number
using System;

class GFG {
    
// This function will return n % d.
// d must be one of: 1, 2, 4, 8, 16, 32, … 
static uint getModulo( uint n, uint d)
{
return ( n & (d-1) );
}     

// Driver code
static public void Main () 
   {
    uint n = 6;
    uint d = 4; /*d must be a power of 2*/

    Console.WriteLine( n + " moduo " + d + 
                " is " + getModulo(n, d));
    
    }
}
// This code is contributed by vt_m.

PHP

<?php
// This function will return n % d.
// d must be one of: 1, 2, 4, 8, 16, 32, … 
function getModulo($n, $d)
{
return ( $n & ($d - 1) );
}     

// Driver Code
$n = 6;

// d must be a power of 2
$d = 4; 
echo $n ," moduo"," ", $d, " is ", 
         " ",getModulo($n, $d);
    
// This code is contributed by vt_m.
?>


References:
http://graphics.stanford.edu/~seander/bithacks.html#ModulusDivisionEasy

Please write comments if you find any bug in the above program/algorithm or other ways to solve the same problem.



My Personal Notes arrow_drop_up

Improved By : vt_m




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