Can you extract bitmask directly from bitfield in C ?-CodePudding

Considering following example:

#include <iostream>

using namespace std;

struct Test
{
    uint8_t A:1;
    uint8_t B:1;
    uint8_t C:1;
    uint8_t D:1;
};

int main()
{
    Test test;
    test.A = 1;
    test.B = 0;
    test.C = 1;
    test.D = 0;
    
    int bitmask = test.A | (test.B << 1) | (test.C << 2) | (test.D << 3);
    
    cout << "Size: " << sizeof(test) << ", bitmask: " << bitmask;

    return 0;
}

I'm assuming that the data in the bitfield is represented as bitmask somehow? I was wondering if there is a way to get a bitmask directly, without having to go through and shift all members. In this case it's not a big deal, but if you have large bitfield it can get pretty tedious.

For example it would be great if I could do something like this:

int bitmask = (int)test;

Of course that doesn't work. Any way to achieve similar robustness?

CodePudding user response：

Assuming you want to convert the entire struct, and there exists an integral type with the same size as the struct:

Test test;
test.A = 1;
test.B = 0;
test.C = 1;
test.D = 0;

cout << (int)std::bit_cast<char>(test) << '\n';

std::bit_cast is a C 20 feature.

char is used here because it has the same size as Test. I'm casting the result to int, because otherwise cout interpretes the resulting number as a character code.

The alternative solution is to memcpy the struct (or a part of it) to an integer:

char ch;
std::memcpy(&ch, &test, 1);
cout << (int)ch << '\n';

Unlike bit_cast, this doesn't require a modern compiler, and allows you to inspect a specific part of the struct (assuming the part is byte-aligned).