为什么std :: bitset <8>变量无法处理11111111？

Question

为什么这个程序显示以下输出？

#include <bitset>
#include <cstdio>
#include <iostream>

int main()
{
    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.
                                                           //Same is the case with b1
    std::cout << "b1 & b2: " << (b1 & b2) << '\n';
    std::cout << "b1 | b2: " << (b1 | b2) << '\n';
    std::cout << "b1 ^ b2: " << (b1 ^ b2) << '\n';
getchar();
return 0;
}

这是输出:

01000000
11000111
b1 & b2: 01000000
b1 | b2: 11000111
b1 ^ b2: 10000111

首先,我认为头文件有问题(我使用的是MinGW)所以我使用MSVCC进行了检查.但它也显示了同样的事情.请帮忙!

Answer 1

你发起了std::bitset一个数字 01100100和11111111.只需查看输出:

01000000
11000111

两者都错了.你正在发起std::bitset一个十进制数11111111它101010011000101011000111的二进制形式.首先是数字01100100以八进制表示:1001000000001000000二进制.并std::bitset采取最重要的八位.

这是正确的代码:

#include <bitset>
#include <iostream>
#include <string>

int main()
{
    std::bitset<8> b1(std::string("01100100")); std::cout<<b1<<std::endl;
    std::bitset<8> b2(std::string("11111111")); std::cout<<b2<<std::endl;
    std::cout << "b1 & b2: " << (b1 & b2) << '\n';
    std::cout << "b1 | b2: " << (b1 | b2) << '\n';
    std::cout << "b1 ^ b2: " << (b1 ^ b2) << '\n';
    return 0;
}

而正确的输出:

01100100
11111111
b1 & b2: 01100100
b1 | b2: 11111111
b1 ^ b2: 10011011