Contrary to the answers here, you DON'T need to worry about encoding if the bytes don't need to be interpreted!
Like you mentioned, your goal is, simply, to "get what bytes the string has been stored in".
(And, of course, to be able to re-construct the string from the bytes.)
For those goals, I honestly do not understand why people keep telling you that you need the encodings. You certainly do NOT need to worry about encodings for this.
Just do this instead:
static byte[] GetBytes(string str)
{
byte[] bytes = new byte[str.Length * sizeof(char)];
System.Buffer.BlockCopy(str.ToCharArray(), 0, bytes, 0, bytes.Length);
return bytes;
}
// Do NOT use on arbitrary bytes; only use on GetBytes's output on the SAME system
static string GetString(byte[] bytes)
{
char[] chars = new char[bytes.Length / sizeof(char)];
System.Buffer.BlockCopy(bytes, 0, chars, 0, bytes.Length);
return new string(chars);
}
As long as your program (or other programs) don't try to interpret the bytes somehow, which you obviously didn't mention you intend to do, then there is nothing wrong with this approach! Worrying about encodings just makes your life more complicated for no real reason.
Additional benefit to this approach: It doesn't matter if the string contains invalid characters, because you can still get the data and reconstruct the original string anyway!
It will be encoded and decoded just the same, because you are just looking at the bytes.
If you used a specific encoding, though, it would've given you trouble with encoding/decoding invalid characters.
I heard LINQ is the new black, so here's my attempt using LINQ:
private static Random random = new Random();
public static string RandomString(int length)
{
const string chars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
return new string(Enumerable.Repeat(chars, length)
.Select(s => s[random.Next(s.Length)]).ToArray());
}
(Note: The use of the Random
class makes this unsuitable for anything security related, such as creating passwords or tokens. Use the RNGCryptoServiceProvider
class if you need a strong random number generator.)
Best Answer
You can use formatting in the strings defined in your resources. When you use
{0}
, as shown inFieldRequired
, the display name will be inserted when possible. Otherwise it will fall back to the property name as shown forMiddleName
.Example:
Resources:
Strings.resx
Strings.nl.resx
Implementation: