Convert array of bits to an array of bytes

Question

I want to convert an array of bits (bool* bitArray) where the values are 1s and 0s into an array of bytes (unsigned char* byteArray) where the values at each index would be one byte. For ex, index 0~7 in bitArray would go into byteArray[1].

How would I go about doing this? Assuming that I already have an array of bits (but the amount would be subject to change based on the incoming data). I am not worried about having it divisible by 8 because I will just add padding at the end of the bitArray to make it divisible by 8.

"in hex" describes a way to display numbers. I don't believe this question involves anything being displayed or in hex. — Drew Dormann
– Drew Dormann, Commented Jun 2, 2022 at 17:30
I'm really missing a bitcast from std::bitset to std:array and back. — Goswin von Brederlow
– Goswin von Brederlow, Commented Jun 2, 2022 at 18:15

fabian · Accepted Answer · 2022-06-02 17:56:45Z

Just just use bit shifts or a lookup array and and combine numbers with 1 bit set each with bitwise or for 8 bits at a time:

int main() {
    bool input[] = {
        false, false, false, true, true, true, false, false, false,
        false, false, false, true, true, true, false, false, false,
        false, false, false, true, true, true, false, false, false,
        false, false, false, true, true, true, false, false, false,
    };

    constexpr auto len = sizeof(input) / sizeof(*input);
    constexpr size_t outLen = ((len % 8 == 0) ? 0 : 1) + len / 8;

    uint8_t out[outLen];

    bool* inPos = input;
    uint8_t* outPos = out;

    size_t remaining = len;

    // output bytes where there are all 8 bits available
    for (; remaining >= 8; remaining -= 8, ++outPos)
    {
        uint8_t value = 0;
        for (size_t i = 0; i != 8; ++i, ++inPos)
        {
            if (*inPos)
            {
                value |= (1 << (7 - i));
            }
        }
        *outPos = value;
    }

    if (remaining != 0)
    {
        // output byte that requires padding
        uint8_t value = 0;
        for (size_t i = 0; i != remaining; ++i, ++inPos)
        {
            if (*inPos)
            {
                value |= (1 << (7 - i));
            }
        }
        *outPos = value;
    }

    for (auto v : out)
    {
        std::cout << static_cast<int>(v) << '\n';
    }

    return 0;
}

The rhs of the |= operator could also be replaced with a lookup in the following array, if you consider this simpler to understand:

constexpr uint8_t Bits[8]
{
    0b1000'0000,
    0b0100'0000,
    0b0010'0000,
    0b0001'0000,
    0b0000'1000,
    0b0000'0100,
    0b0000'0010,
    0b0000'0001,
};

...
value |= Bits[i];
...

Goswin von Brederlow · Accepted Answer · 2022-06-02 19:03:28Z

You should be using std::bitset for an array of bools, or std::vector<bool> if it's dynamically sized. And std::array for the array or again std::vector for dynamic size. I've only done static size below and conversion to and from.

Converting involves a lot of bit shifts and loops for something that should be a memcpy (on little endian or unsigned char types). The compiler output for -O2 is bad. -O3 removes the loop and to_array2 gets interesting. gcc nearly manages to optimize it, clang actually gets it down to movzx eax, word ptr [rdi]: https://godbolt.org/z/4chb8o81e

#include <array>
#include <bitset>
#include <climits>

template <typename T, std::size_t len>
constexpr std::bitset<sizeof(T) * CHAR_BIT * len> from_array(const std::array<T, len> &arr) {
    std::bitset<sizeof(T) * CHAR_BIT * len> res;
    std::size_t pos = 0;
    for (auto x : arr) {
        for(std::size_t i = 0; i < sizeof(T) * CHAR_BIT; ++i) {
            res[pos++] = x & 1;
            x >>= 1;
        }
    }
    return res;
}

template <typename T, std::size_t len>
constexpr std::array<T, (len + sizeof(T) * CHAR_BIT - 1) / (sizeof(T) * CHAR_BIT)> to_array(const std::bitset<len> &bit) {
    std::array<T, (len + sizeof(T) * CHAR_BIT - 1) / (sizeof(T) * CHAR_BIT)> res;
    T mask = 1;
    T t = 0;
    std::size_t pos = 0;
    for (std::size_t i = 0; i < len; ++i) {
        if (bit[i]) t |= mask;
        mask <<= 1;
        if (mask == 0) {
            mask = 1;
            res[pos++] = t;
            t = 0;
        }
    }
    if constexpr (len % (sizeof(T) * CHAR_BIT) != 0) {
        res[pos] = t;
    }
    return res;
}

std::bitset<16> from_array2(const std::array<unsigned char, 2> &arr) {
    return from_array(arr);
}

std::array<unsigned short, 1> to_array2(const std::bitset<16> &bits) {
    return to_array<unsigned short>(bits);
}

#include <iostream>

int main() {
    std::array<unsigned char, 2> arr{0, 255};
    std::bitset bits = from_array(arr);
    std::cout << bits << std::endl;

    std::bitset<16> bits2{0x1234};
    std::array<unsigned short, 1> arr2 = to_array<unsigned short>(bits2);
    std::cout << std::hex << arr2[0] << std::endl;
}

Collectives™ on Stack Overflow

Convert array of bits to an array of bytes

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related