Convert Byte Array to Bit Array?

How would I go about开发者_开发百科 converting a bytearray to a bit array?

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The obvious way; using the constructor that takes a byte array:

BitArray bits = new BitArray(arrayOfBytes);

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It depends on what you mean by "bit array"... If you mean an instance of the BitArray class, Guffa's answer should work fine.

If you actually want an array of bits, in the form of a bool[] for instance, you could do something like that :

byte[] bytes = ...
bool[] bits = bytes.SelectMany(GetBits).ToArray();

...

IEnumerable<bool> GetBits(byte b)
{
    for(int i = 0; i < 8; i++)
    {
        yield return (b & 0x80) != 0;
        b *= 2;
    }
}

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public static byte[] ToByteArray(this BitArray bits)
 {
    int numBytes = bits.Count / 8;
    if (bits.Count % 8 != 0) numBytes++;
    byte[] bytes = new byte[numBytes];
    int byteIndex = 0, bitIndex = 0;
    for (int i = 0; i < bits.Count; i++) {
        if (bits[i])
            bytes[byteIndex] |= (byte)(1 << (7 - bitIndex));
        bitIndex++;
        if (bitIndex == 8) {
            bitIndex = 0;
            byteIndex++;
        }
    }
    return bytes;
}

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You can use BitArray to create a stream of bits from a byte array. Here an example:

string testMessage = "This is a test message";

byte[] messageBytes = Encoding.ASCII.GetBytes(testMessage);

BitArray messageBits = new BitArray(messageBytes);

---

byte number  = 128;
Convert.ToString(number, 2);

=> out: 10000000

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public static byte[] ToByteArray(bool[] byteArray)
{
    return = byteArray
               .Select(
                    (val1, idx1) => new { Index = idx1 / 8, Val = (byte)(val1 ? Math.Pow(2, idx1 % 8) : 0) }
                )
                .GroupBy(gb => gb.Index)
                .Select(val2 => (byte)val2.Sum(s => (byte)s.Val))
                .ToArray();
}