Why is the std::bitset<8> variable unable to handle 11111111?

Question

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';
}

This is the OUTPUT:

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

First, I thought there is something wrong with the header file (I was using MinGW) so I checked using MSVCC. But it too showed the same thing. Please help.

Answer 1

Despite the appearance, the 11111111 is decimal. The binary representation of 11111111₁₀ is 101010011000101011000111₂. Upon construction, std::bitset<8> takes the eight least significant bits of that: 11000111₂.

The first case is similar except the 01100100 is octal (due to the leading zero). The same number expressed in binary is 1001000000001000000₂.

One way to represent a bitset with a value of 11111111₂ is std::bitset<8> b1(0xff).

Alternatively, you can construct a bitset from a binary string:

std::bitset<8> b1(std::string("01100100"));
std::bitset<8> b2(std::string("11111111"));