Binary Serialization of std::bitset
std::bitset
has a to_string()
method for serializing as a char
-based string of 1
s and 0
s. Obviously, this uses a single 8 bit char
for each bit in the bitset, making the serialized representation 8 times longer than necessary.
to_ulong()
method is relevant only when there are less than 32 bits in my bitset. I have hundreds.
I'm not sure I want to use memcpy()
/std::copy()
on the object (address) itself, as that assumes the object is a POD.
The API does not seem to provide a handle to the internal array representation from which I could have taken the address.
I would also like the option to deserialize the bitset from the binary representation.
How can I do this?
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
.
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.
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.
As suggested by guys at gamedev.net, one can try using boost::dynamic_bitset since it allows access to internal representation of bitpacked data.
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.
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);
}
};
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