c# - WS_EX_TOOLWINDOW with TransparencyKey causes Win32Exception

Question

I'm trying to have an empty form window, but using the tool window style. However, calling Show() results in the following exception:

Win32Exception: The parameter is incorrect.

NativeErrorCode: 87

at System.Windows.Forms.Form.UpdateLayered() at System.Windows.Forms.Control.WmCreate(Message& m) at System.Windows.Forms.Control.WndProc(Message& m) at System.Windows.Forms.Form.WmCreate(Message& m) at System.Windows.Forms.Form.WndProc(Message& m) at System.Windows.Forms.NativeWindow.DebuggableCallback(IntPtr hWnd, Int32 msg, IntPtr wparam, IntPtr lparam)

Error code 87 is ERROR_INVALID_PARAMETER.

private class ToolForm : Form {
    public ToolForm() {
        AllowTransparency = true;
        BackColor = System.Drawing.Color.FromArgb(0, 0, 1);
        TransparencyKey = BackColor;
    }

    private const int WS_EX_TOOLWINDOW = 0x00000080;
    protected override CreateParams CreateParams {
        get {
            var cp = base.CreateParams;
            cp.ExStyle = WS_EX_TOOLWINDOW;
            return cp;
        }
    }
}

Edit:

This works:

public class ToolForm : Form {
    public ToolForm() {
        this.FormBorderStyle = System.Windows.Forms.FormBorderStyle.SizableToolWindow;
        this.AllowTransparency = true;
        this.BackColor = Color.FromArgb(0, 0, 1);
        this.TransparencyKey = this.BackColor;
    }
}

Answer 1

First try using an OR-assignment instead of plain assignment:

cp.ExStyle |= WS_EX_TOOLWINDOW;

If that doesn't work, you can try additionally OR'ing some of these related styles:

cp.ExStyle |= ( int )(
  WS_EX_LAYERED |
  WS_EX_TRANSPARENT |
  WS_EX_NOACTIVATE |
  WS_EX_TOOLWINDOW );

The associated values are:

WS_EX_LAYERED = 0x00080000,
WS_EX_NOACTIVATE = 0x08000000,
WS_EX_TOOLWINDOW = 0x00000080,
WS_EX_TRANSPARENT = 0x00000020

The WS_EX_TRANSPARENT flag may possibly allow the transparency you want without requiring the TransparencyKey = BackColor; line.