开发者

Check if class has been instantiated

开发者 https://www.devze.com 2023-01-02 00:49 出处:网络
In my current program I have a main window and a secondary window that pops up when a button is pressed. If the secondary window is currently shown but doesn\'t have focus the button will instead brin

In my current program I have a main window and a secondary window that pops up when a button is pressed. If the secondary window is currently shown but doesn't have focus the button will instead bring it to focus.

At this time I am creating a new instance of the secondary window as the main window loads and simply checking its status with SubWindow.IsDisposed and SubWindow.CanFocus

I have found that if I do not create a new instance at the beginning SubWindow.IsDisposed throws an exception. As long as I'd previously created an instance of SubWindow the check runs fine.

My question- The current version works fine but is there a better way of doing this? It is not a huge concern, but it feels like it'd be better to be able to check for existence without having to gua开发者_如何学编程rantee that it has existed at least once before.


You could do a check on SubWindow to see if it is null. If thats the case then instantiate the SubWindow otherwise it exists.


VoodooChild got me on the right track. (SubWindow == null) returns false though when the window has opened once and then been closed.

Currently using

(SubWindow == null || SubWindow.IsDisposed)

which works for all cases so far.


You can have static counter property in your class. Increment on instantiation, decrement on disposal. That's in general... in your case you better follow VoodooChild's advice.


Implement the second window using a singleton pattern.


public class SecondForm : Form
{
       public static m_myInstance= new SecondForm();
       public static bool m_visible = false;

       public SecondForm ()
       {
              InitializeComponent()               
       }

       public SecondForm Instance()
       {
            return m_myInstance;
       }

      public static void Show()
      {
          ...
      }

} 
0

精彩评论

暂无评论...
验证码 换一张
取 消