宏ut_2pow_remainder

求余数 12%8=4

n%m也能计算出余数,但效率可能比位操作要低一些

/*************************************************************//**
Calculates fast the remainder of n/m when m is a power of two.
@param n    in: numerator
@param m    in: denominator, must be a power of two
@return        the remainder of n/m */
#define ut_2pow_remainder(n, m) ((n) & ((m) - 1))

#include <stdio.h>
#include <stdlib.h>

/*************************************************************//**
Calculates fast the remainder of n/m when m is a power of two.
@param n    in: numerator
@param m    in: denominator, must be a power of two
@return        the remainder of n/m */
#define ut_2pow_remainder(n, m) ((n) & ((m) - 1))

int main(){
    int n=12;
    int m=8;
    int a;
    a=ut_2pow_remainder(n,m);
    printf("%d",a);
    return 0;
}