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;
}

Random number generator only generating one random number

Every time you do new Random() it is initialized using the clock. This means that in a tight loop you get the same value lots of times. You should keep a single Random instance and keep using Next on the sameinstance.

//Function to get random number
private static readonly Random random = new Random();
private static readonly object syncLock = new object();
public static int RandomNumber(int min, int max)
{
    lock(syncLock) { // synchronize
        return random.Next(min, max);
    }
}

---

Edit (see comments): why do we need a lock here?

Basically, Next is going to change the internal state of the Random instance. If we do that at the same time from multiple threads, you could argue "we've just made the outcome even more random", but what we are actually doing is potentially breaking the internal implementation, and we could also start getting the same numbers from different threads, which might be a problem - and might not. The guarantee of what happens internally is the bigger issue, though; since Random does not make any guarantees of thread-safety. Thus there are two valid approaches:

• synchronize so that we don't access it at the same time from different threads
• use different Random instances per thread

either can be fine; but mutating a single instance from multiple callers at the same time is just asking for trouble.

The lock achieves the first (and simpler) of these approaches; however, another approach might be:

private static readonly ThreadLocal<Random> appRandom
     = new ThreadLocal<Random>(() => new Random());

this is then per-thread, so you don't need to synchronize.