判断一个整数是不是二的整数次幂 求某个数的二进制中1的个数

方法：i&(i-1) 

如二进制1100 && 1011 =1000 相当于去掉最右边的1

//求一个整数中二进制1的个数 --普通方法int Numof1(unsigned int i){int count=0;while(i){if(i&1){++count;}i>>=1;}return count;}int GetBitNumOfOne(int n) //http://zh.wikipedia.org/wiki/%E6%B1%89%E6%98%8E%E9%87%8D%E9%87%8F{n = ((0xaaaaaaaa & n) >> 1) + (0x55555555 & n);n = ((0xcccccccc & n) >> 2) + (0x33333333 & n);n = ((0xf0f0f0f0 & n) >> 4) + (0x0f0f0f0f & n);n = ((0xff00ff00 & n) >> 8) + (0x00ff00ff & n);n = ((0xffff0000 & n) >> 16) + (0x0000ffff & n);return n;}//求一个整数中二进制1的个数int Numof1_S(unsigned int i){int count=0;while(i){++count;i&=(i-1);}return count;}//判断一个正整数是不是2的正整数次幂int isPowerBase2(unsigned int i){return (i&(i-1))==0 ? 1 : 0;}#include #include int main(void) {int N=10;printf("%d\n",Numof1(N));printf("%d\n",Numof1_S(N));printf("%d\n",GetBitNumOfOne(N));printf("%d\n",isPowerBase2(1024));return 0;}