Bit Field Struct to int in C++ - c++

Question

I have a struct with same type members in it. I am trying to convert it into uint8_t type. I am able to do that but cannot see the output please tell me where I am going wrong. Also I know there are another ways to do it? I am trying to do it this way because I want to get use to static_cast and reinterpret_cast.
Code is below:
int main()
{
struct xs{
bool x :1 ;
bool y :1;
bool z :1;
uint8_t num :5;
} zs;
uint8_t* P = static_cast<uint8_t*>(static_cast<void*>(&zs));
cout << *P << endl;
return 0;
}

Answer 1

There a lot of problems here:
You seem to believe that x, y, and z will all pack into the single uint_8. This is not the case. "Adjacently declared bit fields of the same type can then be packed by the compiler into a reduced number of words"[1]
"The value of sizeof(bool) is implementation defined and might differ from 1"[2] Therefore your xs will be implementation defined, but certainly not equivilent to sizeof(uint_8)
Because xs is not "similar" to a uint_8 according to the rules defined for C++'s type aliasing the behavior of your reinterpret_cast<uint_8*> is undefined
Finally as has been pointed out by many others the reason that you can't see anything is that whatever implementation defined value is at *P it is likely a control character with no visible representation when treated by cout as a char
A possible workaround to the code you have would be to use these definitions:
constexpr uint8_t X = 0B1000'0000;
constexpr uint8_t Y = 0B0100'0000;
constexpr uint8_t Z = 0B0010'0000;
constexpr uint8_t NUM = 0B0001'1111;
uint8_t zs;
Then given that some value is assigned to zs you can perform these functions to output the former bit fields:
cout << ((zs & X) != 0) << endl;
cout << ((zs & Y) != 0) << endl;
cout << ((zs & Z) != 0) << endl;
cout << (zs & NUM) << endl;
Live Example

Answer 2

it is undefined behavior to access an object through a pointer to a type other than the type of the object. it works with most compilers, but technically your programm is allowed to do whatever it wants.
assuming we are not running into a problem mentioned in 1, my guess is that uint8_t is an alias for char, so cout will put a char onto your console.
you did not initialize the memory you are inspecting, so it can be 0. a char with value zero wont print out anything "observeable" in your console, look at the ascii table. try filling your struct with a 50 for example

Answer 3

There are two things that need to be solved here. First off, as some pointed out, you cannot access your object this way. If you want to construct a uint8_t properly, you would need to read the variables inside your struct and do some bit shifts, something like this:
uint8_t value = 0;
value |= (zs.x ? 1 << 7 : 0);
value |= (zs.y ? 1 << 6 : 0);
value |= (zs.z ? 1 << 5 : 0);
value |= zs.num;
Now, the second problem you are facing is that you're trying to output a number of 8-bits wide. By default, this is interpreted as a 'character' and will display as such. In order to accomplish what you want to do, you can either use a variable with a different length (uint16_t, uint32_t, ...) or use std::to_string.
cout << std::to_string(value) << endl;