移位运算符包括左移"<<"和右移">>"
左移运算符<<:
1.无符号
语法格式:需要移位的数字<<移位的次数n
运算规则:按二进制形式把所有数字向左移动相应的位数,高位移出(舍弃),低位的空位补0。相当于乘以2的n次方
例如:4<<2 ,就是将数字4左移2位
过程:4的二进制形式:00000000 00000000 00000000 00000100;然后把高位2个0移出,其余所有位向左移动2位,低位补0,得到:00000000 00000000 00000000 00010000;十进制数为16,16=4*22。
C++程序测试:
#include <iostream>
#include <bitset>
using namespace std;
int main() {
unsigned short short1 = 4;
bitset<16> bitset1{short1}; // the bitset representation of 4
cout << bitset1 << endl; // 0000000000000100 unsigned short short2 = short1 << 1; // 4 left-shifted by 1 = 8
bitset<16> bitset2{short2};
cout << bitset2 << endl; // 0000000000001000 unsigned short short3 = short1 << 2; // 4 left-shifted by 2 = 16
bitset<16> bitset3{short3};
cout << bitset3 << endl; // 0000000000010000
}
2.有符号
如果你左移有符号的数字,以至于符号位受影响,则结果是不确定的。
C++程序测试:
#include <iostream>
#include <bitset>
using namespace std; int main() {
short short1 = 16384;
bitset<16> bitset1{short2};
cout << bitset1 << endl; // 0100000000000000 short short3 = short1 << 1;
bitset<16> bitset3{short3}; // 16384 left-shifted by 1 = -32768
cout << bitset3 << endl; // 100000000000000 short short4 = short1 << 14;
bitset<16> bitset4{short4}; // 4 left-shifted by 14 = 0
cout << bitset4 << endl; // 000000000000000
}
右移运算符>>:
1.无符号
语法格式:需要移位的数字>>移位的次数n
运算规则:按二进制形式把所有数字向右移动相应的位数,低位移出(舍弃),高位的空位补0。相当于除以2的n次方
例如:4>>2 ,就是将数字4左移2位
过程:4的二进制形式:00000000 00000000 00000000 00000100;然后把低位2个0移出,其余所有位向右移动2位,高位补0,得到:00000000 00000000 00000000 00000001;十进制数为1,1=4÷22。
C++程序测试:
#include <iostream>
#include <bitset>
using namespace std; int main() {
unsigned short short11 = 1024;
bitset<16> bitset11{short11};
cout << bitset11 << endl; // 0000010000000000 unsigned short short12 = short11 >> 1; // 512
bitset<16> bitset12{short12};
cout << bitset12 << endl; // 0000001000000000 unsigned short short13 = short11 >> 10; // 1
bitset<16> bitset13{short13};
cout << bitset13 << endl; // 0000000000000001 unsigned short short14 = short11 >> 11; // 0
bitset<16> bitset14{short14};
cout << bitset14 << endl; // 0000000000000000}
}
2.有符号
语法格式:需要移位的数字>>移位的次数n
运算规则:按二进制形式把所有数字向右移动相应的位数,低位移出(舍弃),正数,高位的空位补0。负数,高位的空位补1.
C++程序测试:
正数:
#include <iostream>
#include <bitset>
using namespace std; int main() {
short short1 = 1024;
bitset<16> bitset1{short1};
cout << bitset1 << endl; // 0000010000000000 short short2 = short1 >> 1; // 512
bitset<16> bitset2{short2};
cout << bitset2 << endl; // 0000001000000000 short short3 = short1 >> 11; // 0
bitset<16> bitset3{short3};
cout << bitset3 << endl; // 0000000000000000
} 负数:
#include <iostream>
#include <bitset>
using namespace std; int main() {
short neg1 = -16;
bitset<16> bn1{neg1};
cout << bn1 << endl; // 1111111111110000 short neg2 = neg1 >> 1; // -8
bitset<16> bn2{neg2};
cout << bn2 << endl; // 1111111111111000 short neg3 = neg1 >> 2; // -4
bitset<16> bn3{neg3};
cout << bn3 << endl; // 1111111111111100 short neg4 = neg1 >> 4; // -1
bitset<16> bn4{neg4};
cout << bn4 << endl; // 1111111111111111 short neg5 = neg1 >> 5; // -1
bitset<16> bn5{neg5};
cout << bn5 << endl; // 1111111111111111
}
参考:
http://baike.baidu.com/link?url=N6rRNKFdHoXOqfzDT1vEf1ASgGReSgTEBoBkIzCSCNzKE1gNe8ViDNIWFRbqGqXvMahOPD5o8eifyl3Fhzezwa
https://msdn.microsoft.com/zh-cn/library/336xbhcz.aspx