C++移位运算符详解

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

 

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