Binary Serialization of std::bitset

Question

std::bitset has a to_string() method for serializing as a char-based string of 1s and 0s. Obviously, this us

Answer 1

this might help you, it's a little example of various serialization types. I added bitset and raw bit values, that can be used like the below.

(all examples at https://github.com/goblinhack/simple-c-plus-plus-serializer)

class BitsetClass {
public:
    std::bitset<1> a;
    std::bitset<2> b;
    std::bitset<3> c;

    unsigned int d:1; // need c++20 for default initializers for bitfields
    unsigned int e:2;
    unsigned int f:3;
    BitsetClass(void) { d = 0; e = 0; f = 0; }

    friend std::ostream& operator<<(std::ostream &out,
                                    Bits<const class BitsetClass & > const m
    {
        out << bits(my.t.a);
        out << bits(my.t.b);
        out << bits(my.t.c);

        std::bitset<6> s(my.t.d | my.t.e << 1 | my.t.f << 3);
        out << bits(s);

        return (out);
    }

    friend std::istream& operator>>(std::istream &in,
                                    Bits<class BitsetClass &> my)
    {
        std::bitset<1> a;
        in >> bits(a);
        my.t.a = a;

        in >> bits(my.t.b);
        in >> bits(my.t.c);
        std::bitset<6> s;
        in >> bits(s);

        unsigned long raw_bits = static_cast<unsigned long>(s.to_ulong());
        my.t.d = raw_bits & 0b000001;
        my.t.e = (raw_bits & 0b000110) >> 1;
        my.t.f = (raw_bits & 0b111000) >> 3;

        return (in);
    }
};

Answer 2

edit: The following does not work as intended. Appearently, "binary format" actually means "ASCII representation of binary".

---

You should be able to write them to a std::ostream using operator<<. It says here:

[Bitsets] can also be directly inserted and extracted from streams in binary format.

Answer 3

This is a possible approach based on explicit creation of an std::vector<unsigned char> by reading/writing one bit at a time...

template<size_t N>
std::vector<unsigned char> bitset_to_bytes(const std::bitset<N>& bs)
{
    std::vector<unsigned char> result((N + 7) >> 3);
    for (int j=0; j<int(N); j++)
        result[j>>3] |= (bs[j] << (j & 7));
    return result;
}

template<size_t N>
std::bitset<N> bitset_from_bytes(const std::vector<unsigned char>& buf)
{
    assert(buf.size() == ((N + 7) >> 3));
    std::bitset<N> result;
    for (int j=0; j<int(N); j++)
        result[j] = ((buf[j>>3] >> (j & 7)) & 1);
    return result;
}

Note that to call the de-serialization template function bitset_from_bytes the bitset size N must be specified in the function call, for example

std::bitset<N> bs1;
...
std::vector<unsigned char> buffer = bitset_to_bytes(bs1);
...
std::bitset<N> bs2 = bitset_from_bytes<N>(buffer);

If you really care about speed one solution that would gain something would be doing a loop unrolling so that the packing is done for example one byte at a time, but even better is just to write your own bitset implementation that doesn't hide the internal binary representation instead of using std::bitset.

Answer 4

As suggested by guys at gamedev.net, one can try using boost::dynamic_bitset since it allows access to internal representation of bitpacked data.

Answer 5

Answering my own question for completeness.

Apparently, there is no simple and portable way of doing this.

For simplicity (though not efficiency), I ended up using to_string, and then creating consecutive 32-bit bitsets from all 32-bit chunks of the string (and the remainder*), and using to_ulong on each of these to collect the bits into a binary buffer.
This approach leaves the bit-twiddling to the STL itself, though it is probably not the most efficient way to do this.

* Note that since std::bitset is templated on the total bit-count, the remainder bitset needs to use some simple template meta-programming arithmetic.

Answer 6

I can't see an obvious way other than converting to a string and doing your own serialization of the string that groups chunks of 8 characters into a single serialized byte.

EDIT: Better is to just iterate over all the bits with operator[] and manually serialize it.