问题描述
这是为什么程序显示以下输出？
Why is this program showing the following output ?
#include <bitset>
...
{
std::bitset<8> b1(01100100); std::cout<<b1<<std::endl;
std::bitset<8> b2(11111111); std::cout<<b2<<std::endl; //see, this variable
//has been assigned
//the value 11111111
//whereas, during
//execution, it takes
//the value 11000111
std::cout << "b1 & b2: " << (b1 & b2) << '\n';
std::cout << "b1 | b2: " << (b1 | b2) << '\n';
std::cout << "b1 ^ b2: " << (b1 ^ b2) << '\n';
}
这是输出：
01000000
11000111
b1 & b2: 01000000
b1 | b2: 11000111
b1 ^ b2: 10000111
首先，我觉得有什么不对的头文件（我是用MinGW的），所以我
使用MSVCC检查。但它也显示出同样的事情。请帮助。
First, I thought there is something wrong with the header file (I was using MinGW) so Ichecked using MSVCC. But it too showed the same thing. Please help.
推荐答案
尽管外观上， 11111111 是小数。的 11111111 10 是 101010011000101011000111 。
Despite the appearance, the 11111111 is decimal. The binary representation of 11111111 is 101010011000101011000111. Upon construction, std::bitset<8> takes the eight least significant bits of that: 11000111.
第一种情况是，除了 01100100 是八进制（由于前导零）相似。二进制pssed相同数量的前$ P $是 1001000000001000000 。
The first case is similar except the 01100100 is octal (due to the leading zero). The same number expressed in binary is 1001000000001000000.
重新present一个bitset与 11111111 是和std :: bitset℃的价值的一种方式8是氢; B1（0xFF的）。
One way to represent a bitset with a value of 11111111 is std::bitset<8> b1(0xff).
另外，你可以从一个二进制字符串构造一个bitset：
Alternatively, you can construct a bitset from a binary string:
std::bitset<8> b1(std::string("01100100"));
std::bitset<8> b2(std::string("11111111"));
这篇关于为什么的std :: bitset的＆LT; 8'变量无法处理11111111？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！