How do you convert Byte Array to Hexadecimal String, and vice versa?

public static string ByteArrayToString(byte[] ba)
{
  StringBuilder hex = new StringBuilder(ba.Length * 2);
  foreach (byte b in ba)
    hex.AppendFormat("{0:x2}", b);
  return hex.ToString();
}

or:

public static string ByteArrayToString(byte[] ba)
{
  string hex = BitConverter.ToString(ba);
  return hex.Replace("-","");
}

There are even more variants of doing it, for example here.

The reverse conversion would go like this:

public static byte[] StringToByteArray(String hex)
{
  int NumberChars = hex.Length;
  byte[] bytes = new byte[NumberChars / 2];
  for (int i = 0; i < NumberChars; i += 2)
    bytes[i / 2] = Convert.ToByte(hex.Substring(i, 2), 16);
  return bytes;
}

---

Edit: you can improve performance for long strings by using a single pass parser, like so:

public static byte[] StringToByteArray(String hex)
{
  int NumberChars = hex.Length/2;
  byte[] bytes = new byte[NumberChars];
  using (var sr = new StringReader(hex))
  {
    for (int i = 0; i < NumberChars; i++)
      bytes[i] = 
        Convert.ToByte(new string(new char[2]{(char)sr.Read(), (char)sr.Read()}), 16);
  }
  return bytes;
}

Two mashups which folds the two nibble operations into one.

Probably pretty efficient version:
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-with-bit-hacking version:
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))) );
}

And reverse:
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;
}