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

C++

Java

Python3

C#

PHP

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