How would I go about converting a bytearray to a bit array?
The obvious way; using the constructor that takes a byte array:
BitArray bits = new BitArray(arrayOfBytes);
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;
}
byte number = 128;
Convert.ToString(number, 2);
=> out: 10000000
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;
}
}
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);
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();
}