在 C# 中,是否有一种方法可以对整个字节数组进行右/左移位操作(并在最后一位添加一个字节以防止丢失)?
我知道这听起来像是一个奇怪的请求,但我仍然想知道是否可能以及如何开始实现它。
在 C# 中,是否有一种方法可以对整个字节数组进行右/左移位操作(并在最后一位添加一个字节以防止丢失)?
我知道这听起来像是一个奇怪的请求,但我仍然想知道是否可能以及如何开始实现它。
仅供娱乐。移动和旋转字节数组中的字节。(不是位移)
向左移位,填充零:
mybytes.Skip(1).Concat(new byte[] { 0 }).ToArray();
向右移位,填充零:
(new byte[] {0}).Concat(mybytes.Take(mybytes.Length - 1)).ToArray();
向左旋转:
mybytes.Skip(1).Concat(mybytes.Take(1)).ToArray();
向右旋转:
mybytes.Skip(mbytes.Length - 1).Concat(mbytes.Take(mbytes.Length - 1)).ToArray();
Skip
函数。 - Ben Voigt是的,您可以。请看我写的以下方法:
/// <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;
}
workRegister
是有符号的,当你不想要符号传播时?) - Ben Voigt看起来你正在对大量位进行位运算,并将它们存储在字节数组中。考虑使用BitArray类和BitVector32结构。根据你对位的操作,你可以创建一个像这样的类。请注意,移位的时间复杂度为O(1),而不是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();
}
}
BitArray
没有提供任何移位运算符或方法。 - Ben VoigtBitArray
中,你可以在一个整数中存储指向起始索引的指针。移位操作将是 - 在末尾添加一个元素并将起始指针向右移动一个位置。这将以 O(1) 的时间复杂度运行。如果你有一系列位操作而不仅仅是这个移位操作,使用 BitArray
可能是一个好主意。 - George Mamaladzepublic 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;
}
左移:
for (int i = byteArray.Length - 1; i >= 0; i--) byteArray[i] = (byte)((byteArray[i] << 1) | ((i == 0) ? 0 : byteArray[i - 1] >> 7));
右移:
for (int i = 0; i <= byteArray.Length - 1; i++) byteArray[i] = (byte)((byteArray[i] >> 1) | ((i == byteArray.Length - 1) ? 0 : byteArray[i + 1] << 7));
Linq方式:
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();
}
}
我认为没有内置的方法。我实现了你在下面描述的左移操作(假设是小端)。它不像使用x86汇编语言那样优雅(带进位指令),但与使用C语言相当接近。
或者,你几乎可以使用BigInteger结构体(.NET 4及以上版本),它有一个以字节数组为参数的构造函数和一个ToByteArray方法。但它的左移操作会符号扩展高字节,右移操作会截断。因此,你需要对两者进行补偿,才能得到你所描述的精确行为。
// 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);
}
这两个函数将按指定的数量移动字节数组中的位,根据需要将它们移动到相邻的字节中。可选地,它们可以将数组的位从一端包装到另一端。请注意,它们创建一个新的数组,但代码可以轻松更改以修改传递的“数组”...
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;
}
}
ʀ[]
数组不存在,但它会反复尝试使用它。我试图建议编辑,但“建议的编辑队列已满”。 - Daniel
BigInteger
。 - leppieArray.Copy()
是否适合我正在做的事情。无论如何。 - Bitterblue