这是一篇很棒的帖子。我喜欢瓦利德的解决方案。我还没有通过帕特里奇的测试，但似乎很快。我还需要反向过程，将十六进制字符串转换为字节数组，因此我将其作为Waleed解决方案的反向来编写。不确定它是否比托马拉克的原始解决方案更快。同样，我也没有通过帕特里奇的测试运行相反的过程。

private byte[] HexStringToByteArray(string hexString)
{
    int hexStringLength = hexString.Length;
    byte[] b = new byte[hexStringLength / 2];
    for (int i = 0; i < hexStringLength; i += 2)
    {
        int topChar = (hexString[i] > 0x40 ? hexString[i] - 0x37 : hexString[i] - 0x30) << 4;
        int bottomChar = hexString[i + 1] > 0x40 ? hexString[i + 1] - 0x37 : hexString[i + 1] - 0x30;
        b[i / 2] = Convert.ToByte(topChar + bottomChar);
    }
    return b;
}

Waleed Eissa代码的逆函数（十六进制字符串到字节数组）：

    public static byte[] HexToBytes(this string hexString)        
    {
        byte[] b = new byte[hexString.Length / 2];            
        char c;
        for (int i = 0; i < hexString.Length / 2; i++)
        {
            c = hexString[i * 2];
            b[i] = (byte)((c < 0x40 ? c - 0x30 : (c < 0x47 ? c - 0x37 : c - 0x57)) << 4);
            c = hexString[i * 2 + 1];
            b[i] += (byte)(c < 0x40 ? c - 0x30 : (c < 0x47 ? c - 0x37 : c - 0x57));
        }

        return b;
    }

Waleed Eissa功能，支持小写：

    public static string BytesToHex(this byte[] barray, bool toLowerCase = true)
    {
        byte addByte = 0x37;
        if (toLowerCase) addByte = 0x57;
        char[] c = new char[barray.Length * 2];
        byte b;
        for (int i = 0; i < barray.Length; ++i)
        {
            b = ((byte)(barray[i] >> 4));
            c[i * 2] = (char)(b > 9 ? b + addByte : b + 0x30);
            b = ((byte)(barray[i] & 0xF));
            c[i * 2 + 1] = (char)(b > 9 ? b + addByte : b + 0x30);
        }

        return new string(c);
    }

在编写加密代码时，通常避免依赖数据的分支和表查找，以确保运行时不依赖于数据，因为依赖数据的计时可能会导致侧通道攻击。

它也很快。

static string ByteToHexBitFiddle(byte[] bytes)
{
    char[] c = new char[bytes.Length * 2];
    int b;
    for (int i = 0; i < bytes.Length; i++) {
        b = bytes[i] >> 4;
        c[i * 2] = (char)(55 + b + (((b-10)>>31)&-7));
        b = bytes[i] & 0xF;
        c[i * 2 + 1] = (char)(55 + b + (((b-10)>>31)&-7));
    }
    return new string(c);
}

Ph'nglui mglw'naph Cthulhu R'lieh wgah'nagl fhtagn公司

你们这些进入这里的人，放弃一切希望吧

一个奇怪的拨弄解释：

bytes[i]>>4提取字节的高位半字节bytes[i]&0xF提取字节的低位半字节b-10对于值b<10，为<0，将变为十进制数字对于值b>10，为>=0，这将成为从a到F的字母。在有符号32位整数上使用i>>31可以提取符号，这得益于符号扩展。当i<0时为-1，当i>=0时为0。结合2）和3），表明（b-10）>>31将是字母0，数字-1。看看字母的大小写，最后一个被加数变为0，b在10到15的范围内。我们希望将其映射到A（65）到F（70），这意味着添加55（'A'-10）。看看数字的情况，我们希望调整最后一个被加数，使其将b从范围0到9映射到范围0（48）到9（57）。这意味着它需要变为-7（'0'-55）。现在我们可以乘以7。但由于-1由所有位表示为1，因此我们可以改用&-7，因为（0&-7）==0和（-1&-7）==-7。

进一步考虑：

我没有使用第二个循环变量来索引c，因为测量表明从I计算它更便宜。正好使用i<bytes.Length作为循环的上限允许JITter消除对bytes[i]的边界检查，所以我选择了这个变量。将b设为int允许不必要的从和到字节的转换。

扩展BigInteger方法（Gregory Morse在上面提到过）。我不能评论效率，它使用System.Linq.Reverse（），但它很小而且内置。

        // To hex
        byte[] bytes = System.Text.Encoding.UTF8.GetBytes("Test String!£");
        string hexString = new System.Numerics.BigInteger(bytes.Reverse().ToArray()).ToString("x2");

        // From hex
        byte[] fromHexBytes = System.Numerics.BigInteger.Parse(hexString, System.Globalization.NumberStyles.HexNumber).ToByteArray().Reverse().ToArray();

        // Unit test
        CollectionAssert.AreEqual(bytes, fromHexBytes);

这里不想赘述很多答案，但我发现了一个十六进制字符串解析器的相当优化（比公认的好4.5倍）、简单的实现。首先，我的测试输出（第一批是我的实现）：

Give me that string:
04c63f7842740c77e545bb0b2ade90b384f119f6ab57b680b7aa575a2f40939f

Time to parse 100,000 times: 50.4192 ms
Result as base64: BMY/eEJ0DHflRbsLKt6Qs4TxGfarV7aAt6pXWi9Ak58=
BitConverter'd: 04-C6-3F-78-42-74-0C-77-E5-45-BB-0B-2A-DE-90-B3-84-F1-19-F6-AB-5
7-B6-80-B7-AA-57-5A-2F-40-93-9F

Accepted answer: (StringToByteArray)
Time to parse 100000 times: 233.1264ms
Result as base64: BMY/eEJ0DHflRbsLKt6Qs4TxGfarV7aAt6pXWi9Ak58=
BitConverter'd: 04-C6-3F-78-42-74-0C-77-E5-45-BB-0B-2A-DE-90-B3-84-F1-19-F6-AB-5
7-B6-80-B7-AA-57-5A-2F-40-93-9F

With Mono's implementation:
Time to parse 100000 times: 777.2544ms
Result as base64: BMY/eEJ0DHflRbsLKt6Qs4TxGfarV7aAt6pXWi9Ak58=
BitConverter'd: 04-C6-3F-78-42-74-0C-77-E5-45-BB-0B-2A-DE-90-B3-84-F1-19-F6-AB-5
7-B6-80-B7-AA-57-5A-2F-40-93-9F

With SoapHexBinary:
Time to parse 100000 times: 845.1456ms
Result as base64: BMY/eEJ0DHflRbsLKt6Qs4TxGfarV7aAt6pXWi9Ak58=
BitConverter'd: 04-C6-3F-78-42-74-0C-77-E5-45-BB-0B-2A-DE-90-B3-84-F1-19-F6-AB-5
7-B6-80-B7-AA-57-5A-2F-40-93-9F

base64和“BitConverter'd”行用于测试正确性。请注意，它们是相等的。

实施：

public static byte[] ToByteArrayFromHex(string hexString)
{
  if (hexString.Length % 2 != 0) throw new ArgumentException("String must have an even length");
  var array = new byte[hexString.Length / 2];
  for (int i = 0; i < hexString.Length; i += 2)
  {
    array[i/2] = ByteFromTwoChars(hexString[i], hexString[i + 1]);
  }
  return array;
}

private static byte ByteFromTwoChars(char p, char p_2)
{
  byte ret;
  if (p <= '9' && p >= '0')
  {
    ret = (byte) ((p - '0') << 4);
  }
  else if (p <= 'f' && p >= 'a')
  {
    ret = (byte) ((p - 'a' + 10) << 4);
  }
  else if (p <= 'F' && p >= 'A')
  {
    ret = (byte) ((p - 'A' + 10) << 4);
  } else throw new ArgumentException("Char is not a hex digit: " + p,"p");

  if (p_2 <= '9' && p_2 >= '0')
  {
    ret |= (byte) ((p_2 - '0'));
  }
  else if (p_2 <= 'f' && p_2 >= 'a')
  {
    ret |= (byte) ((p_2 - 'a' + 10));
  }
  else if (p_2 <= 'F' && p_2 >= 'A')
  {
    ret |= (byte) ((p_2 - 'A' + 10));
  } else throw new ArgumentException("Char is not a hex digit: " + p_2, "p_2");

  return ret;
}

我尝试了一些不安全的东西，并将（显然是冗余的）字符移动到另一个方法来蚕食if序列，但这是最快的。

（我承认这回答了一半的问题。我觉得字符串->字节[]转换不足，而字节[]->字符串角度似乎被很好地覆盖了。因此，这个答案。）

我将参加这个比特拨弄比赛，因为我有一个同样使用比特拨弄来解码十六进制的答案。请注意，使用字符数组可能会更快，因为调用StringBuilder方法也需要时间。

public static String ToHex (byte[] data)
{
    int dataLength = data.Length;
    // pre-create the stringbuilder using the length of the data * 2, precisely enough
    StringBuilder sb = new StringBuilder (dataLength * 2);
    for (int i = 0; i < dataLength; i++) {
        int b = data [i];

        // check using calculation over bits to see if first tuple is a letter
        // isLetter is zero if it is a digit, 1 if it is a letter
        int isLetter = (b >> 7) & ((b >> 6) | (b >> 5)) & 1;

        // calculate the code using a multiplication to make up the difference between
        // a digit character and an alphanumerical character
        int code = '0' + ((b >> 4) & 0xF) + isLetter * ('A' - '9' - 1);
        // now append the result, after casting the code point to a character
        sb.Append ((Char)code);

        // do the same with the lower (less significant) tuple
        isLetter = (b >> 3) & ((b >> 2) | (b >> 1)) & 1;
        code = '0' + (b & 0xF) + isLetter * ('A' - '9' - 1);
        sb.Append ((Char)code);
    }
    return sb.ToString ();
}

public static byte[] FromHex (String hex)
{

    // pre-create the array
    int resultLength = hex.Length / 2;
    byte[] result = new byte[resultLength];
    // set validity = 0 (0 = valid, anything else is not valid)
    int validity = 0;
    int c, isLetter, value, validDigitStruct, validDigit, validLetterStruct, validLetter;
    for (int i = 0, hexOffset = 0; i < resultLength; i++, hexOffset += 2) {
        c = hex [hexOffset];

        // check using calculation over bits to see if first char is a letter
        // isLetter is zero if it is a digit, 1 if it is a letter (upper & lowercase)
        isLetter = (c >> 6) & 1;

        // calculate the tuple value using a multiplication to make up the difference between
        // a digit character and an alphanumerical character
        // minus 1 for the fact that the letters are not zero based
        value = ((c & 0xF) + isLetter * (-1 + 10)) << 4;

        // check validity of all the other bits
        validity |= c >> 7; // changed to >>, maybe not OK, use UInt?

        validDigitStruct = (c & 0x30) ^ 0x30;
        validDigit = ((c & 0x8) >> 3) * (c & 0x6);
        validity |= (isLetter ^ 1) * (validDigitStruct | validDigit);

        validLetterStruct = c & 0x18;
        validLetter = (((c - 1) & 0x4) >> 2) * ((c - 1) & 0x2);
        validity |= isLetter * (validLetterStruct | validLetter);

        // do the same with the lower (less significant) tuple
        c = hex [hexOffset + 1];
        isLetter = (c >> 6) & 1;
        value ^= (c & 0xF) + isLetter * (-1 + 10);
        result [i] = (byte)value;

        // check validity of all the other bits
        validity |= c >> 7; // changed to >>, maybe not OK, use UInt?

        validDigitStruct = (c & 0x30) ^ 0x30;
        validDigit = ((c & 0x8) >> 3) * (c & 0x6);
        validity |= (isLetter ^ 1) * (validDigitStruct | validDigit);

        validLetterStruct = c & 0x18;
        validLetter = (((c - 1) & 0x4) >> 2) * ((c - 1) & 0x2);
        validity |= isLetter * (validLetterStruct | validLetter);
    }

    if (validity != 0) {
        throw new ArgumentException ("Hexadecimal encoding incorrect for input " + hex);
    }

    return result;
}

从Java代码转换而来。