tags:

views:

256

answers:

1

If I know that a certain key has been pressed (eg Key.D3), and that the Shift key is also down (Keyboard.IsKeyDown(Key.LeftShift) || Keyboard.IsKeyDown(Key.RightShift)), how can I find out what character that refers to (eg, '#' on US keyboard, UK pound sign on UK keyboard, etc)?

Put another way, how can I find out, programatically, that Shift + Key.D3 produces '#' (it wouldn't on a non-US keyboard).

+5  A: 

If you want to determine what character you will get from a given key with given modifiers, you should use the user32 ToAscii function. Or ToAsciiEx if you want to use a keyboard layout other then the current one.

using System.Runtime.InteropServices;
public static class User32Interop
{
  public static char ToAscii(Keys key, Keys modifiers)
  {
    var outputBuilder = new StringBuilder(2);
    int result = ToAscii((uint)key, 0, GetKeyState(modifiers),
                         outputBuilder, 0);
    if (result == 1)
      return outputBuilder[0];
    else
      throw new Exception("Invalid key");
  }

  private const byte HighBit = 0x80;
  private static byte[] GetKeyState(Keys modifiers)
  {
    var keyState = new byte[256];
    foreach (Keys key in Enum.GetValues(typeof(Keys)))
    {
      if ((modifiers & key) == key)
      {
        keyState[(int)key] = HighBit;
      }
    }
    return keyState;
  }

  [DllImport("user32.dll")]
  private static extern int ToAscii(uint uVirtKey, uint uScanCode,
                                    byte[] lpKeyState,
                                    [Out] StringBuilder lpChar,
                                    uint uFlags);
}

You can now use it like this:

char c = User32Interop.ToAscii(Keys.D3, Keys.ShiftKey); // = '#'

If you need more than one modifier, just or them. Keys.ShiftKey | Keys.AltKey

Samuel
Yes, but WHY? (Don't answer that). How can I find out, programatically, that Shift + Key.D3 produces '#' (it wouldn't on a non-US keyboard).
Darren Oster
Oh, perhaps you should refine your question to make that point clear.
Samuel
Found a solution, let me change answer
Samuel
That looks like a winner! Thanks.
Darren Oster
Annoying that it's such a difficult solution, though. Oh well...
Darren Oster
Well, I'd imagine it's not a very used function and was not a priority for the WinForms team to wrap. At least you can always PInvoke Win32 code.
Samuel