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("-", "")

扩展方法（免责声明：完全未经测试的代码，BTW…）：

public static class ByteExtensions
{
    public static string ToHexString(this byte[] ba)
    {
        StringBuilder hex = new StringBuilder(ba.Length * 2);

        foreach (byte b in ba)
        {
            hex.AppendFormat("{0:x2}", b);
        }
        return hex.ToString();
    }
}

使用Tomalak的三种解决方案之一（最后一种是字符串上的扩展方法）。

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);
}

就速度而言，这似乎比这里的任何东西都要好：

  public static string ToHexString(byte[] data) {
    byte b;
    int i, j, k;
    int l = data.Length;
    char[] r = new char[l * 2];
    for (i = 0, j = 0; i < l; ++i) {
      b = data[i];
      k = b >> 4;
      r[j++] = (char)(k > 9 ? k + 0x37 : k + 0x30);
      k = b & 15;
      r[j++] = (char)(k > 9 ? k + 0x37 : k + 0x30);
    }
    return new string(r);
  }

两个mashup，将两个半字节操作合并为一个。

可能非常有效的版本：

public static string ByteArrayToString2(byte[] ba)
{
    char[] c = new char[ba.Length * 2];
    for( int i = 0; i < ba.Length * 2; ++i)
    {
        byte b = (byte)((ba[i>>1] >> 4*((i&1)^1)) & 0xF);
        c[i] = (char)(55 + b + (((b-10)>>31)&-7));
    }
    return new string( c );
}

Decadent linq与比特黑客版本：

public static string ByteArrayToString(byte[] ba)
{
    return string.Concat( ba.SelectMany( b => new int[] { b >> 4, b & 0xF }).Select( b => (char)(55 + b + (((b-10)>>31)&-7))) );
}

并反转：

public static byte[] HexStringToByteArray( string s )
{
    byte[] ab = new byte[s.Length>>1];
    for( int i = 0; i < s.Length; i++ )
    {
        int b = s[i];
        b = (b - '0') + ((('9' - b)>>31)&-7);
        ab[i>>1] |= (byte)(b << 4*((i&1)^1));
    }
    return ab;
}

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。）