Develop Reference

Is there a function to do circular bitshift for a byte array in C#?

I can't seem to find if there's a built-in way to do a circular bitshift of a byte array, what C's ROL and ROR used to do with a single byte?
Let me explain, say, I have an array (in binary):
[0] = 11001110
[1] = 01000100
[2] = 10100001
and then if I want to do ROL_Array(1 bit) or move bits 1 bit to the left, I'd get:
[0] = 10011100
[1] = 10001001
[2] = 01000011
or, if I want to do ROR_Array(2 bits) or move bits 2 bits to the right, I'd get:
[0] = 00110011
[1] = 01010001
[2] = 10101000

This is not as simple as you'd think. Here's a quick version before this thread gets closed:
public static byte[] ROL_ByteArray(byte[] arr, int nShift)
{
//Performs bitwise circular shift of 'arr' by 'nShift' bits to the left
//RETURN:
// = Result
byte[] resArr = new byte[arr.Length];
if(arr.Length > 0)
{
int nByteShift = nShift / (sizeof(byte) * 8); //Adjusted after #dasblinkenlight's correction
int nBitShift = nShift % (sizeof(byte) * 8);
if (nByteShift >= arr.Length)
nByteShift %= arr.Length;
int s = arr.Length - 1;
int d = s - nByteShift;
for (int nCnt = 0; nCnt < arr.Length; nCnt++, d--, s--)
{
while (d < 0)
d += arr.Length;
while (s < 0)
s += arr.Length;
byte byteS = arr[s];
resArr[d] |= (byte)(byteS << nBitShift);
resArr[d > 0 ? d - 1 : resArr.Length - 1] |= (byte)(byteS >> (sizeof(byte) * 8 - nBitShift));
}
}
return resArr;
}
and here's a test:
byte[] arr = new byte[] {
Convert.ToByte("11001110", 2),
Convert.ToByte("01000100", 2),
Convert.ToByte("10100001", 2),
};
byte[] arr2 = Auth.ROL_ByteArray(arr, 1);
string sss = "";
for (int i = 0; i < arr2.Length; i++)
sss += Convert.ToString(arr2[i], 2) + ", ";
Debug.WriteLine(sss);

c# - left shift an entire byte array

In C#, is there a way to right/left shift an entire byte array (and subsequently adding a byte to a particular side for the last bit isn't lost)?
I know this sounds like a weird request, but I'd still like to know if its possible and/or how to begin doing it.

Just for grins. shifting and rotating bytes in a byte array. (not bitshifting)
shift left, zero fill:
mybytes.Skip(1).Concat(new byte[] { 0 }).ToArray();
shift right, zero fill:
(new byte[] {0}).Concat(mybytes.Take(mybytes.Length - 1)).ToArray();
rotate left:
mybytes.Skip(1).Concat(mybytes.Take(1)).ToArray();
rotate right:
mybytes.Skip(mbytes.Length - 1).Concat(mbytes.Take(mbytes.Length - 1)).ToArray();

Yes, you can. See the following methods I wrote:
/// <summary>
/// Rotates the bits in an array of bytes to the left.
/// </summary>
/// <param name="bytes">The byte array to rotate.</param>
public static void RotateLeft(byte[] bytes)
{
bool carryFlag = ShiftLeft(bytes);
if (carryFlag == true)
{
bytes[bytes.Length - 1] = (byte)(bytes[bytes.Length - 1] | 0x01);
}
}
/// <summary>
/// Rotates the bits in an array of bytes to the right.
/// </summary>
/// <param name="bytes">The byte array to rotate.</param>
public static void RotateRight(byte[] bytes)
{
bool carryFlag = ShiftRight(bytes);
if (carryFlag == true)
{
bytes[0] = (byte)(bytes[0] | 0x80);
}
}
/// <summary>
/// Shifts the bits in an array of bytes to the left.
/// </summary>
/// <param name="bytes">The byte array to shift.</param>
public static bool ShiftLeft(byte[] bytes)
{
bool leftMostCarryFlag = false;
// Iterate through the elements of the array from left to right.
for (int index = 0; index < bytes.Length; index++)
{
// If the leftmost bit of the current byte is 1 then we have a carry.
bool carryFlag = (bytes[index] & 0x80) > 0;
if (index > 0)
{
if (carryFlag == true)
{
// Apply the carry to the rightmost bit of the current bytes neighbor to the left.
bytes[index - 1] = (byte)(bytes[index - 1] | 0x01);
}
}
else
{
leftMostCarryFlag = carryFlag;
}
bytes[index] = (byte)(bytes[index] << 1);
}
return leftMostCarryFlag;
}
/// <summary>
/// Shifts the bits in an array of bytes to the right.
/// </summary>
/// <param name="bytes">The byte array to shift.</param>
public static bool ShiftRight(byte[] bytes)
{
bool rightMostCarryFlag = false;
int rightEnd = bytes.Length - 1;
// Iterate through the elements of the array right to left.
for (int index = rightEnd; index >= 0; index--)
{
// If the rightmost bit of the current byte is 1 then we have a carry.
bool carryFlag = (bytes[index] & 0x01) > 0;
if (index < rightEnd)
{
if (carryFlag == true)
{
// Apply the carry to the leftmost bit of the current bytes neighbor to the right.
bytes[index + 1] = (byte)(bytes[index + 1] | 0x80);
}
}
else
{
rightMostCarryFlag = carryFlag;
}
bytes[index] = (byte)(bytes[index] >> 1);
}
return rightMostCarryFlag;
}

It seems you are performing bit operations on large amount of bits storing them in a byte array. Consider using BitArray class and BitVector32 Structure. Depending on what you are doing with bits you can create a class like this. Note that shifting works in O(1) instead of O(n).
public class BitRing : IEnumerable<bool>
{
private readonly BitArray m_InnerBitArray;
private int m_StarIndex;
public BitRing(byte[] bytes)
{
m_InnerBitArray = new BitArray(bytes);
m_StarIndex = 0;
}
public void ShiftLeft()
{
m_StarIndex++;
}
public void ShiftRight()
{
m_StarIndex--;
}
public bool this[int i]
{
get
{
int index = GetIndex(i);
return m_InnerBitArray[index];
}
set
{
int index = GetIndex(i);
m_InnerBitArray[index] = value;
}
}
private int GetIndex(int i)
{
return i - m_StarIndex%m_InnerBitArray.Count;
}
public IEnumerator<bool> GetEnumerator()
{
for (int i = m_StarIndex; i < m_InnerBitArray.Count; i++)
{
yield return m_InnerBitArray[i];
}
for (int i = 0; i < m_StarIndex; i++)
{
yield return m_InnerBitArray[i];
}
}
IEnumerator IEnumerable.GetEnumerator()
{
return GetEnumerator();
}
}

I've given it some more thought and realized that this probably fits the question better:
public static void Main()
{
byte[] bytes = new byte[] { 0xFF, 0x01, 0x80, 0x81 };
Stack<bool> bitStack = CreateBitStack(bytes);
ShiftLeftExpand(bitStack, 1);
byte[] newBytes = CreateByteArray(bitStack);
}
public static void ShiftLeftExpand(Stack<bool> bitStack, int count)
{
while (count-- > 0)
{
bitStack.Push(false);
}
}
public static Stack<bool> CreateBitStack(byte[] bytes)
{
Stack<bool> bitStack = new Stack<bool>(bytes.Length * 8);
for (int bytePosition = 0; bytePosition < bytes.Length; bytePosition++)
{
for (int bitPosition = 7; bitPosition >= 0; bitPosition--)
{
int bitMask = 0x01 << bitPosition;
bitStack.Push((bytes[bytePosition] & bitMask) > 0);
}
}
return bitStack;
}
public static byte[] CreateByteArray(Stack<bool> bitStack)
{
int newArrayLength = (int)Math.Ceiling(bitStack.Count / 8.0);
byte[] bytes = new byte[newArrayLength];
int bitCounter = 0;
while (bitStack.Count > 0)
{
bool? bitValue = bitStack.Pop();
int bitPosition = bitCounter % 8;
int bytePosition = newArrayLength - 1 - bitCounter / 8;
if (bitValue == true)
{
bytes[bytePosition] = (byte)(bytes[bytePosition] | (0x01 << bitPosition));
}
bitCounter++;
}
return bytes;
}
A similar technique can be applied to perform the right shift.

Linq way:
static class Shifter
{
public static byte[] ShiftLeft(this byte[] array, int n)
{
var a = array.Select(x => (byte)(x >> 8 - n % 8)).Concat(new byte[(7 + n) / 8]).Select((x, i) => new Tuple<int, byte>(i - (n % 8 == 0 ? 0 : 1), x));
var b = array.Select(x => (byte)(x << n % 8)).Concat(new byte[n / 8]).Select((x, i) => new Tuple<int, byte>(i, x));
return (from x in a
join y in b on x.Item1 equals y.Item1 into yy
from y in yy.DefaultIfEmpty()
select (byte)(x.Item2 | (y == null ? 0 : y.Item2))).ToArray();
}
public static byte[] ShiftRight(this byte[] array, int n)
{
return (new byte[n/8]).Concat(ShiftLeft(array, (8 - (n%8))%8)).ToArray();
}
}

I don't think there's a built-in way. I implemented the shift-left operation you described below (assuming little endian). It's not quite as elegant as you can do with x86 assembly (shift with carry instructions), but pretty close to what you could do with C.
Alternately, you can almost use the BigInteger struct (.NET 4 and above) which has a constructor that takes a byte array and a ToByteArray method. But its shift left operation sign-extends the high byte and its shift right operation truncates. So you'd need to compensate for both to get the exact behavior you described.
// Left-shifts a byte array in place. Assumes little-endian. Throws on overflow.
static public void ShiftByteArrayLeft(byte[] array)
{
if (array == null)
throw new ArgumentNullException("array");
if (array[array.Length - 1] >= 0x80)
throw new OverflowException();
// move left-to-right, left-shifting each byte
for (int i = array.Length - 1; i >= 1; --i)
{
// left-shift current byte
array[i] <<= 1;
// carry the bit from the next/right byte if needed
if (array[i - 1] >= 0x80)
++array[i];
}
// finally shift the left-shift the right-most byte
array[0] <<= 1;
}
// Left-shifts a byte array in place. Assumes little-endian. Grows array as needed.
static public void ShiftByteArrayLeftAutoGrow(ref byte[] array)
{
if (array == null)
throw new ArgumentNullException("array");
if (array[array.Length - 1] >= 0x80)
{
// allocate a bigger array and do the left-shift on it
byte[] oldArray = array;
array = new byte[oldArray.Length + 1];
Array.Copy(oldArray, 0, array, 0, oldArray.Length);
}
ShiftByteArrayLeft(array);
}

Shift left:
for (int i = byteArray.Length - 1; i >= 0; i--) byteArray[i] = (byte)((byteArray[i] << 1) | ((i == 0) ? 0 : byteArray[i - 1] >> 7));
Shift right:
for (int i = 0; i <= byteArray.Length - 1; i++) byteArray[i] = (byte)((byteArray[i] >> 1) | ((i == byteArray.Length - 1) ? 0 : byteArray[i + 1] << 7));

These two functions will shift the bits in an array of bytes the specified amount, shifting them into neighboring bytes as needed. Optionally, they can wrap the bits from one end of the array to the other. Note that they create a new array, but the code can be easily changed to modify the passed 'array' instead...
public static byte[] ShiftRight(byte[] array, int shift, bool wrap = false) {
if(shift < 0) {
throw new ArgumentOutOfRangeException("shift");
} else if(shift == 0) {
return (byte[])array.Clone();
} else {
if(wrap) shift %= (array.Length * 8);
if(shift >= (array.Length * 8)) return new byte[array.Length];
var offset = (shift / 8);
shift %= 8;
var ʀ = new byte[array.Length];
for(var ɪ = 0; ɪ < ʀ.Length; ɪ++) {
var indexL_ɪ = (ɪ + offset);
var indexH_ɪ = (ɪ + offset + 1);
if(wrap) {
if(indexL_ɪ >= array.Length) indexL_ɪ -= array.Length;
if(indexH_ɪ >= array.Length) indexH_ɪ -= array.Length;
}
if(indexL_ɪ < array.Length) ʀ[ɪ] = (byte)(array[indexL_ɪ] >> shift);
if(indexH_ɪ < array.Length) ʀ[ɪ] |= (byte)(array[indexH_ɪ] << (8 - shift));
}
return ʀ;
}
}
public static byte[] ShiftLeft(byte[] array, int shift, bool wrap = false) {
if(shift < 0) {
throw new ArgumentOutOfRangeException("shift");
} else if(shift == 0) {
return (byte[])array.Clone();
} else {
if(wrap) shift %= (array.Length * 8);
if(shift >= (array.Length * 8)) return new byte[array.Length];
var offset = (shift / 8);
shift %= 8;
for(var ɪ = 0; ɪ < ʀ.Length; ɪ++) {
var indexL_ɪ = (ɪ - offset - 1);
var indexH_ɪ = (ɪ - offset);
if(wrap) {
if(indexL_ɪ < 0) indexL_ɪ += array.Length;
if(indexH_ɪ < 0) indexH_ɪ += array.Length;
}
if(indexL_ɪ >= 0) ʀ[ɪ] = (byte)(array[indexL_ɪ] >> (8 - shift));
if(indexH_ɪ >= 0) ʀ[ɪ] |= (byte)(array[indexH_ɪ] << shift);
}
return ʀ;
}
}

Convert Byte Array to Bit Array?

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);

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;
}
}

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;
}

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

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();
}

Bit-shifting a byte array by N bits

Hello quick question regarding bit shifting
I have a value in HEX: new byte[] { 0x56, 0xAF };
which is 0101 0110 1010 1111
I want to the first N bits, for example 12.
Then I must right-shift off the lowest 4 bits (16 - 12) to get 0000 0101 0110 1010 (1386 dec).
I can't wrap my head around it and make it scalable for n bits.

Sometime ago i coded these two functions, the first one shifts an byte[] a specified amount of bits to the left, the second does the same to the right:
Left Shift:
public byte[] ShiftLeft(byte[] value, int bitcount)
{
byte[] temp = new byte[value.Length];
if (bitcount >= 8)
{
Array.Copy(value, bitcount / 8, temp, 0, temp.Length - (bitcount / 8));
}
else
{
Array.Copy(value, temp, temp.Length);
}
if (bitcount % 8 != 0)
{
for (int i = 0; i < temp.Length; i++)
{
temp[i] <<= bitcount % 8;
if (i < temp.Length - 1)
{
temp[i] |= (byte)(temp[i + 1] >> 8 - bitcount % 8);
}
}
}
return temp;
}
Right Shift:
public byte[] ShiftRight(byte[] value, int bitcount)
{
byte[] temp = new byte[value.Length];
if (bitcount >= 8)
{
Array.Copy(value, 0, temp, bitcount / 8, temp.Length - (bitcount / 8));
}
else
{
Array.Copy(value, temp, temp.Length);
}
if (bitcount % 8 != 0)
{
for (int i = temp.Length - 1; i >= 0; i--)
{
temp[i] >>= bitcount % 8;
if (i > 0)
{
temp[i] |= (byte)(temp[i - 1] << 8 - bitcount % 8);
}
}
}
return temp;
}
If you need further explanation please comment on this, i will then edit my post for clarification...

You can use a BitArray and then easily copy each bit to the right, starting from the right.
http://msdn.microsoft.com/en-us/library/system.collections.bitarray_methods.aspx

you want something like...
var HEX = new byte[] {0x56, 0xAF};
var bits = new BitArray(HEX);
int bitstoShiftRight = 4;
for (int i = 0; i < bits.Length; i++)
{
bits[i] = i < (bits.Length - bitstoShiftRight) ? bits[i + bitstoShiftRight] : false;
}
bits.CopyTo(HEX, 0);

If you have k total bits, and you want the "first" (as in most significant) n bits, you can simply right shift k-n times. The last k-n bits will be removed, by sort of "falling" off the end, and the first n will be moved to the least significant side.

Answering using C-like notation, assuming bits_in_byte is the number of bits in a byte determined elsewhere:
int remove_bits_count= HEX.count*bits_in_byte - bits_to_keep;
int remove_bits_in_byte_count= remove_bits_count % bits_in_byte;
if (remove_bits_count > 0)
{
for (int iteration= 0; iteration<min(HEX.count, (bits_to_keep + bits_in_byte - 1)/bits_in_byte); ++iteration)
{
int write_index= HEX.count - iteration - 1;
int read_index_lo= write_index - remove_bits_count/bits_in_byte;
if (read_index_lo>=0)
{
int read_index_hi= read_index_lo - (remove_bits_count + bits_in_byte - 1)/bits_in_byte;
HEX[write_index]=
(HEX[read_index_lo] >> remove_bits_in_byte_count) |
(HEX[read_index_hi] << (bits_in_byte - remove_bits_in_byte_count));
}
else
{
HEX[write_index]= 0;
}
}
}
Assuming you are overwriting the original array, you basically take every byte you write to and figure out the bytes that it would get its shifted bits from. You go from the end of the array to the front to ensure you never overwrite data you will need to read.

Converting a range into a bit array

I'm writing a time-critical piece of code in C# that requires me to convert two unsigned integers that define an inclusive range into a bit field. Ex:
uint x1 = 3;
uint x2 = 9;
//defines the range [3-9]
// 98 7654 3
//must be converted to: 0000 0011 1111 1000
It may help to visualize the bits in reverse order
The maximum value for this range is a parameter given at run-time which we'll call max_val. Therefore, the bit field variable ought to be defined as a UInt32 array with size equal to max_val/32:
UInt32 MAX_DIV_32 = max_val / 32;
UInt32[] bitArray = new UInt32[MAX_DIV_32];
Given a range defined by the variables x1 and x2, what is the fastest way to perform this conversion?

Try this. Calculate the range of array items that must be filled with all ones and do this by iterating over this range. Finally set the items at both borders.
Int32 startIndex = x1 >> 5;
Int32 endIndex = x2 >> 5;
bitArray[startIndex] = UInt32.MaxValue << (x1 & 31);
for (Int32 i = startIndex + 1; i <= endIndex; i++)
{
bitArray[i] = UInt32.MaxValue;
}
bitArray[endIndex] &= UInt32.MaxValue >> (31 - (x2 & 31));
May be the code is not 100% correct, but the idea should work.
Just tested it and found three bugs. The calculation at start index required a mod 32 and at end index the 32 must be 31 and a logical and instead of a assignment to handle the case of start and end index being the same. Should be quite fast.
Just benchmarked it with equal distribution of x1 and x2 over the array.
Intel Core 2 Duo E8400 3.0 GHz, MS VirtualPC with Server 2003 R2 on Windows XP host.
Array length [bits] 320 160 64
Performance [executions/s] 33 million 43 million 54 million
One more optimazation x % 32 == x & 31 but I am unable to meassure a performance gain. Because of only 10.000.000 iterations in my test the fluctuations are quite high. And I am running in VirtualPC making the situation even more unpredictable.

My solution for setting a whole range of bits in a BitArray to true or false:
public static BitArray SetRange(BitArray bitArray, Int32 offset, Int32 length, Boolean value)
{
Int32[] ints = new Int32[(bitArray.Count >> 5) + 1];
bitArray.CopyTo(ints, 0);
var firstInt = offset >> 5;
var lastInt = (offset + length) >> 5;
Int32 mask = 0;
if (value)
{
// set first and last int
mask = (-1 << (offset & 31));
if (lastInt != firstInt)
ints[lastInt] |= ~(-1 << ((offset + length) & 31));
else
mask &= ~(-1 << ((offset + length) & 31));
ints[firstInt] |= mask;
// set all ints in between
for (Int32 i = firstInt + 1; i < lastInt; i++)
ints[i] = -1;
}
else
{
// set first and last int
mask = ~(-1 << (offset & 31));
if (lastInt != firstInt)
ints[lastInt] &= -1 << ((offset + length) & 31);
else
mask |= -1 << ((offset + length) & 31);
ints[firstInt] &= mask;
// set all ints in between
for (Int32 i = firstInt + 1; i < lastInt; i++)
ints[i] = 0;
}
return new BitArray(ints) { Length = bitArray.Length };
}

You could try:
UInt32 x1 = 3;
UInt32 x2 = 9;
UInt32 newInteger = (UInt32)(Math.Pow(2, x2 + 1) - 1) &
~(UInt32)(Math.Pow(2, x1)-1);

Is there a reason not to use the System.Collections.BitArray class instead of a UInt32[]? Otherwise, I'd try something like this:
int minIndex = (int)x1/32;
int maxIndex = (int)x2/32;
// first handle the all zero regions and the all one region (if any)
for (int i = 0; i < minIndex; i++) {
bitArray[i] = 0;
}
for (int i = minIndex + 1; i < maxIndex; i++) {
bitArray[i] = UInt32.MaxValue; // set to all 1s
}
for (int i = maxIndex + 1; i < MAX_DIV_32; i++) {
bitArray[i] = 0;
}
// now handle the tricky parts
uint maxBits = (2u << ((int)x2 - 32 * maxIndex)) - 1; // set to 1s up to max
uint minBits = ~((1u << ((int)x1 - 32 * minIndex)) - 1); // set to 1s after min
if (minIndex == maxIndex) {
bitArray[minIndex] = maxBits & minBits;
}
else {
bitArray[minIndex] = minBits;
bitArray[maxIndex] = maxBits;
}

I was bored enough to try doing it with a char array and using Convert.ToUInt32(string, int) to convert to a uint from base 2.
uint Range(int l, int h)
{
char[] buffer = new char[h];
for (int i = 0; i < buffer.Length; i++)
{
buffer[i] = i < h - l ? '1' : '0';
}
return Convert.ToUInt32(new string(buffer), 2);
}
A simple benchmark shows that my method is about 5% faster than Angrey Jim's (even if you replace second Pow with a bit shift.)
It is probably the easiest to convert to producing a uint array if the upper bound is too big to fit into a single int. It's a little cryptic but I believe it works.
uint[] Range(int l, int h)
{
char[] buffer = new char[h];
for (int i = 0; i < buffer.Length; i++)
{
buffer[i] = i < h - l ? '1' : '0';
}
int bitsInUInt = sizeof(uint) * 8;
int numNeededUInts = (int)Math.Ceiling((decimal)buffer.Length /
(decimal)bitsInUInt);
uint[] uints = new uint[numNeededUInts];
for (int j = uints.Length - 1, s = buffer.Length - bitsInUInt;
j >= 0 && s >= 0;
j--, s -= bitsInUInt)
{
uints[j] = Convert.ToUInt32(new string(buffer, s, bitsInUInt), 2);
}
int remainder = buffer.Length % bitsInUInt;
if (remainder > 0)
{
uints[0] = Convert.ToUInt32(new string(buffer, 0, remainder), 2);
}
return uints;
}

Try this:
uint x1 = 3;
uint x2 = 9;
int cbToShift = x2 - x1; // 6
int nResult = ((1 << cbToShift) - 1) << x1;
/*
(1<<6)-1 gives you 63 = 111111, then you shift it on 3 bits left
*/