位运算

位运算是针对二进制的运算，也就是说进行位运算是需要先把数据转换成二进制。位运算总共有6种运算：按位取反（~）、与（&）、或（|）、异或（^）、左移（<<）、右移（>>）。运算规律如下图所示：

下面是有关位运算的题：

//二进制中1的个数int GetNumber1(int n)//除法，没有办法处理负数{int count = 0;while(n){if(n%2 != 0){count++;}n /= 2;}return count;}/*int GetNumber2(int n)//右移1位{int count = 0;while(n){if(n & 1)//如果n是负数，左边补1，陷入死循环{count++;}n >>= 1;}return count;}*/int GetNumber2(int n){int count = 0;unsigned int flag = 1;while(flag){if(n & flag){count++;}flag <<= 1;//将1一直左移，而不是将n右移，第一次n&0001 第二次n&0010}return count;}//整数除以2和右移1位在数学上是等价的，但是除法的效率比移位运算要低的多int GetNumber3(int n){int count = 0;while(n){count++;n &= (n-1);//把二进制n中最右边的1去掉}return count;}

对常用的位运算进行总结

//位运算实现加减乘除#include<stdio.h>#include<math.h>//不用加减乘除号求两个数之和int Sum(int num1,int num2){int tmp1 = 0;int tmp2 = 0;do{tmp1 = num1 ^ num2;tmp2 = (num1 & num2) << 1;num1 = tmp1;num2 = tmp2;}while(tmp2 != 0);return tmp1;}//不用加减乘除号求两个数之差int Sub(int num1,int num2){num2 = -num2;return Sum(num1,num2);}//不用加减乘除号求两个数之积int Multi(int num1,int num2){int sum = 0;if(num1 < 0 && num2 < 0){for(int i = 0; i < -num2; ++i){sum = Sum(sum,-num1);}return sum;}if(num1 > 0 && num2 < 0){int tmp = num1;num1 = num2;num2 = tmp;}for(int i = 0; i < num2; ++i){sum = Sum(sum,num1);}return sum;}//不用加减乘除号求两个数之商bool Div(int num1,int num2,int*p){*p = 0;if(num2 == 0){return false;}else if(abs(num1) < abs(num2) || num1 == 0){*p = 0;return true;}else{if(num1 > 0 && num2 > 0){while(num1 > num2){(*p)++;num1 = Sub(num1,num2);}}else if(num1 < 0 || num2 < 0){num1 = abs(num1);num2 = abs(num2);while(num1 > num2){(*p)++;num1 = Sub(num1,num2);}(*p) = -(*p);}else{num1 = abs(num1);num2 = abs(num2);while(num1 > num2){(*p)++;num1 = Sub(num1,num2);}}}return true;}