Convert Byte Array into Bitset

Question

I have a byte array generated by a random number generator. I want to put this into the STL bitset.

Unfortunately, it looks like Bitset only supports the following constructors:

1. A string of 1's and 0's like "10101011"
2. An unsigned long. (my byte array will be longer)

The only solution I can think of now is to read the byte array bit by bit and make a string of 1's and 0's. Does anyone have a more efficient solution?

Answer 1

Something like this?

#include <bitset>
#include <climits>

template<size_t numBytes>
std::bitset<numBytes * CHAR_BIT> bytesToBitset(uint8_t *data)
{
    std::bitset<numBytes * CHAR_BIT> b;

    for(int i = 0; i < numBytes; ++i)
    {
        uint8_t cur = data[i];
        int offset = i * CHAR_BIT;

        for(int bit = 0; bit < CHAR_BIT; ++bit)
        {
            b[offset] = cur & 1;
            ++offset;   // Move to next bit in b
            cur >>= 1;  // Move to next bit in array
        }
    }

    return b;
}

And an example usage:

int main()
{
    std::array<uint8_t, 4> bytes = { 0xDE, 0xAD, 0xBE, 0xEF };
    auto bits = bytesToBitset<bytes.size()>(bytes.data());
    std::cout << bits << std::endl;
}

Answer 2

I have spent a lot of time by writing a reverse function (bitset -> byte/char array). There it is:

    bitset<SIZE> data = ...

    // bitset to char array
    char current = 0;
    int offset = 0;
    for (int i = 0; i < SIZE; ++i) {
        if (data[i]) { // if bit is true
            current |= (char)(int)pow(2, i - offset * CHAR_BIT); // set that bit to true in current masked value
        } // otherwise let it to be false
        if ((i + 1) % CHAR_BIT == 0) { // every 8 bits
            buf[offset++] = current; // save masked value to buffer & raise offset of buffer
            current = 0; // clear masked value
        }
    }
    
    // now we have the result in "buf" (final size of contents in buffer is "offset")

Answer 3

There's a 3rd constructor for bitset<> - it takes no parameters and sets all the bits to 0. I think you'll need to use that then walk through the array calling set() for each bit in the byte array that's a 1.

A bit brute-force, but it'll work. There will be a bit of complexity to convert the byte-index and bit offset within each byte to a bitset index, but it's nothing a little bit of thought (and maybe a run through under the debugger) won't solve. I think it's most likely simpler and more efficient than trying to run the array through a string conversion or a stream.

Answer 4

Here is my implementation using template meta-programming.
Loops are done in the compile-time.
I took @strager version, modified it in order to prepare for TMP:

• changed order of iteration (so that I could make recursion from it);
• reduced number of used variables.

Modified version with loops in a run-time:

template <size_t nOfBytes>
void bytesToBitsetRunTimeOptimized(uint8_t* arr, std::bitset<nOfBytes * CHAR_BIT>& result) {
  for(int i = nOfBytes - 1; i >= 0; --i) {
    for(int bit = 0; bit < CHAR_BIT; ++bit) {
      result[i * CHAR_BIT + bit] = ((arr[i] >> bit) & 1);
    }
  }
}

TMP version based on it:

template<size_t nOfBytes, int I, int BIT> struct LoopOnBIT {
  static inline void bytesToBitset(uint8_t* arr, std::bitset<nOfBytes * CHAR_BIT>& result) {
    result[I * CHAR_BIT + BIT] = ((arr[I] >> BIT) & 1);
    LoopOnBIT<nOfBytes, I, BIT+1>::bytesToBitset(arr, result);
  }
};
// stop case for LoopOnBIT
template<size_t nOfBytes, int I> struct LoopOnBIT<nOfBytes, I, CHAR_BIT> {
  static inline void bytesToBitset(uint8_t* arr, std::bitset<nOfBytes * CHAR_BIT>& result) { }
};

template<size_t nOfBytes, int I> struct LoopOnI {
  static inline void bytesToBitset(uint8_t* arr, std::bitset<nOfBytes * CHAR_BIT>& result) {
    LoopOnBIT<nOfBytes, I, 0>::bytesToBitset(arr, result);
    LoopOnI<nOfBytes, I-1>::bytesToBitset(arr, result);
  }
};
// stop case for LoopOnI
template<size_t nOfBytes> struct LoopOnI<nOfBytes, -1> {
  static inline void bytesToBitset(uint8_t* arr, std::bitset<nOfBytes * CHAR_BIT>& result) { }
};

template <size_t nOfBytes>
void bytesToBitset(uint8_t* arr, std::bitset<nOfBytes * CHAR_BIT>& result) {
  LoopOnI<nOfBytes, nOfBytes - 1>::bytesToBitset(arr, result);
}

client code:

uint8_t arr[]={0x6A};
  std::bitset<8> b; 
  bytesToBitset<1>(arr,b);

Answer 5

Well, let's be honest, I was bored and started to think there had to be a slightly faster way than setting each bit.

template<int numBytes>
std::bitset<numBytes * CHARBIT bytesToBitset(byte *data)
{
    std::bitset<numBytes * CHAR_BIT> b = *data;

    for(int i = 1; i < numBytes; ++i)
    {
        b <<= CHAR_BIT;  // Move to next bit in array
        b |= data[i];    // Set the lowest CHAR_BIT bits
    }

    return b;
}

This is indeed slightly faster, at least as long as the byte array is smaller than 30 elements (depending on your optimization-flags passed to compiler). Larger array than that and the time used by shifting the bitset makes setting each bit faster.

Answer 6

you can initialize the bitset from a stream. I can't remember how to wrangle a byte[] into a stream, but...

from http://www.sgi.com/tech/stl/bitset.html

  bitset<12> x;

  cout << "Enter a 12-bit bitset in binary: " << flush;
  if (cin >> x) {
    cout << "x =        " << x << endl;
    cout << "As ulong:  " << x.to_ulong() << endl;
    cout << "And with mask: " << (x & mask) << endl;
    cout << "Or with mask:  " << (x | mask) << endl;
  }