How to put bit sequence into bytes (C/C++) - c++

Question

I have a couple of integers, for example (in binary represetation):
00001000, 01111111, 10000000, 00000001
and I need to put them in sequence to array of bytes(chars), without the leading zeros, like so:
10001111 11110000 0001000
I understand that it is must be done by bit shifting with <<,>> and using binary or |. But I can't find the correct algorithm, can you suggest the best approach?
The integers I need to put there are unsigned long long ints, so the length of one can be anywhere from 1 bit to 8 bytes (64 bits).

Answer 1

You could use a std::bitset:
#include <bitset>
#include <iostream>
int main() {
unsigned i = 242122534;
std::bitset<sizeof(i) * 8> bits;
bits = i;
std::cout << bits.to_string() << "\n";
}

Answer 2

There are doubtless other ways of doing it, but I would probably go with the simplest:
std::vector<unsigned char> integers; // Has your list of bytes
integers.push_back(0x02);
integers.push_back(0xFF);
integers.push_back(0x00);
integers.push_back(0x10);
integers.push_back(0x01);
std::string str; // Will have your resulting string
for(unsigned int i=0; i < integers.size(); i++)
for(int j=0; j<8; j++)
str += ((integers[i]<<j) & 0x80 ? "1" : "0");
std::cout << str << "\n";
size_t begin = str.find("1");
if(begin > 0) str.erase(0,begin);
std::cout << str << "\n";
I wrote this up before you mentioned that you were using long ints or whatnot, but that doesn't actually change very much of this. The mask needs to change, and the j loop variable, but otherwise the above should work.

Answer 3

Convert them to strings, then erase all leading zeros:
#include <iostream>
#include <sstream>
#include <string>
#include <cstdint>
std::string to_bin(uint64_t v)
{
std::stringstream ss;
for(size_t x = 0; x < 64; ++x)
{
if(v & 0x8000000000000000)
ss << "1";
else
ss << "0";
v <<= 1;
}
return ss.str();
}
void trim_right(std::string& in)
{
size_t non_zero = in.find_first_not_of("0");
if(std::string::npos != non_zero)
in.erase(in.begin(), in.begin() + non_zero);
else
{
// no 1 in data set, what to do?
in = "<no data>";
}
}
int main()
{
uint64_t v1 = 437148234;
uint64_t v2 = 1;
uint64_t v3 = 0;
std::string v1s = to_bin(v1);
std::string v2s = to_bin(v2);
std::string v3s = to_bin(v3);
trim_right(v1s);
trim_right(v2s);
trim_right(v3s);
std::cout << v1s << "\n"
<< v2s << "\n"
<< v3s << "\n";
return 0;
}

Answer 4

A simple approach would be having the "current byte" (acc in the following), the associated number of used bits in it (bitcount) and a vector of fully processed bytes (output):
int acc = 0;
int bitcount = 0;
std::vector<unsigned char> output;
void writeBits(int size, unsigned long long x)
{
while (size > 0)
{
// sz = How many bit we're about to copy
int sz = size;
// max avail space in acc
if (sz > 8 - bitcount) sz = 8 - bitcount;
// get the bits
acc |= ((x >> (size - sz)) << (8 - bitcount - sz));
// zero them off in x
x &= (1 << (size - sz)) - 1;
// acc got bigger and x got smaller
bitcount += sz;
size -= sz;
if (bitcount == 8)
{
// got a full byte!
output.push_back(acc);
acc = bitcount = 0;
}
}
}
void writeNumber(unsigned long long x)
{
// How big is it?
int size = 0;
while (size < 64 && x >= (1ULL << size))
size++;
writeBits(size, x);
}
Note that at the end of the processing you should check if there is any bit still in the accumulator (bitcount > 0) and you should flush them in that case by doing a output.push_back(acc);.
Note also that if speed is an issue then probably using a bigger accumulator is a good idea (however the output will depend on machine endianness) and also that discovering how many bits are used in a number can be made much faster than a linear search in C++ (for example x86 has a special machine language instruction BSR dedicated to this).