我今天遇到了同样的问题，我遇到了以下代码：

private static string ByteArrayToHex(byte[] barray)
{
    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 + 0x37 : b + 0x30);
        b = ((byte)(barray[i] & 0xF));
        c[i * 2 + 1] = (char)(b > 9 ? b + 0x37 : b + 0x30);
    }
    return new string(c);
}

来源：论坛帖子byte[]数组到十六进制字符串（见PZahra的帖子）。我稍微修改了一下代码，删除了0x前缀。

我对代码进行了一些性能测试，它几乎比使用BitConverter.ToString（）快八倍（根据patridge的帖子，速度最快）。

为什么要让它变得复杂？这在Visual Studio 2008中很简单：

C#:

string hex = BitConverter.ToString(YourByteArray).Replace("-", "");

VB:

Dim hex As String = BitConverter.ToString(YourByteArray).Replace("-", "")

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。

另一种方法是使用stackaloc来降低GC内存压力：

static string ByteToHexBitFiddle(byte[] bytes)
{
        var c = stackalloc char[bytes.Length * 2 + 1];
        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));
        }
        c[bytes.Length * 2 ] = '\0';
        return new string(c);
}

安全版本：

public static class HexHelper
{
    [System.Diagnostics.Contracts.Pure]
    public static string ToHex(this byte[] value)
    {
        if (value == null)
            throw new ArgumentNullException("value");

        const string hexAlphabet = @"0123456789ABCDEF";

        var chars = new char[checked(value.Length * 2)];
        unchecked
        {
            for (int i = 0; i < value.Length; i++)
            {
                chars[i * 2] = hexAlphabet[value[i] >> 4];
                chars[i * 2 + 1] = hexAlphabet[value[i] & 0xF];
            }
        }
        return new string(chars);
    }

    [System.Diagnostics.Contracts.Pure]
    public static byte[] FromHex(this string value)
    {
        if (value == null)
            throw new ArgumentNullException("value");
        if (value.Length % 2 != 0)
            throw new ArgumentException("Hexadecimal value length must be even.", "value");

        unchecked
        {
            byte[] result = new byte[value.Length / 2];
            for (int i = 0; i < result.Length; i++)
            {
                // 0(48) - 9(57) -> 0 - 9
                // A(65) - F(70) -> 10 - 15
                int b = value[i * 2]; // High 4 bits.
                int val = ((b - '0') + ((('9' - b) >> 31) & -7)) << 4;
                b = value[i * 2 + 1]; // Low 4 bits.
                val += (b - '0') + ((('9' - b) >> 31) & -7);
                result[i] = checked((byte)val);
            }
            return result;
        }
    }
}

不安全版本适用于那些喜欢性能且不怕不安全的人。ToHex快35%，FromHex快10%。

public static class HexUnsafeHelper
{
    [System.Diagnostics.Contracts.Pure]
    public static unsafe string ToHex(this byte[] value)
    {
        if (value == null)
            throw new ArgumentNullException("value");

        const string alphabet = @"0123456789ABCDEF";

        string result = new string(' ', checked(value.Length * 2));
        fixed (char* alphabetPtr = alphabet)
        fixed (char* resultPtr = result)
        {
            char* ptr = resultPtr;
            unchecked
            {
                for (int i = 0; i < value.Length; i++)
                {
                    *ptr++ = *(alphabetPtr + (value[i] >> 4));
                    *ptr++ = *(alphabetPtr + (value[i] & 0xF));
                }
            }
        }
        return result;
    }

    [System.Diagnostics.Contracts.Pure]
    public static unsafe byte[] FromHex(this string value)
    {
        if (value == null)
            throw new ArgumentNullException("value");
        if (value.Length % 2 != 0)
            throw new ArgumentException("Hexadecimal value length must be even.", "value");

        unchecked
        {
            byte[] result = new byte[value.Length / 2];
            fixed (char* valuePtr = value)
            {
                char* valPtr = valuePtr;
                for (int i = 0; i < result.Length; i++)
                {
                    // 0(48) - 9(57) -> 0 - 9
                    // A(65) - F(70) -> 10 - 15
                    int b = *valPtr++; // High 4 bits.
                    int val = ((b - '0') + ((('9' - b) >> 31) & -7)) << 4;
                    b = *valPtr++; // Low 4 bits.
                    val += (b - '0') + ((('9' - b) >> 31) & -7);
                    result[i] = checked((byte)val);
                }
            }
            return result;
        }
    }
}

顺便提一下对于每次调用的转换函数错误时初始化字母表的基准测试，字母表必须是常量（对于字符串）或静态只读（对于字符[]）。然后，基于字母表的字节[]到字符串的转换变得和字节操作版本一样快。

当然，测试必须在Release中编译（带有优化），并关闭调试选项“抑制JIT优化”（如果代码必须可调试，则“仅启用我的代码”也是如此）。

如果你想得到wcoenen报告的“4倍速度增长”，那么如果不明显：用hex[i]+hex[i+1]替换hex.Substring（i，2）

您还可以再进一步，通过在两个地方使用i++来消除i+=2。