# 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.

```#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 program to test above function */
int main()
{
unsigned int n = 6;
unsigned int d = 4; /*d must be a power of 2*/
printf("%u moduo %u is %u", n, d, getModulo(n, d));

getchar();
return 0;
}
```

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

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