Show function

[Xavier PACOTTE]

Hello,

I would like to add some actions to show function of a form with Visual
Stuio 2005.
Is it possible?
Is it possible to create a new Show function which inherit of old it?

Best Regards.

 

[Marc Gravell]

Perhaps overriding OnLoad() might be more appropriate; what are you
trying to do?

Marc

 

[Moty Michaely]

Hello,

I would like to add some actions to show function of a form with Visual
Stuio 2005.
Is it possible?
Is it possible to create a new Show function which inherit of old it?

Best Regards.

Dear Xavier,

You can override the OnShown method or use the Shown event.

What exactly you want to achieve?

Moty

 

[Miroslav Stampar]

public class FormEx : Form
{
public new void Show()
{
//do your pre-stuff
base.Show();
//do your post-stuff
}

public new void Show(IWin32Window owner)
{
//do your pre-stuff
base.Show(owner);
//do your post-stuff
}
}

Xavier PACOTTE je napisao/la:

 

[Xavier PACOTTE]

thank you for this answer.
it is ok.

public class FormEx : Form
{
public new void Show()
{
//do your pre-stuff
base.Show();
//do your post-stuff
}

public new void Show(IWin32Window owner)
{
//do your pre-stuff
base.Show(owner);
//do your post-stuff
}
}

Xavier PACOTTE je napisao/la:

 

[Marc Gravell]

Warning - this will only work if the caller knows about FormEx; if the
caller only knows about Form, then the original Show will be called,
and your code will not be executed.

As such, this is *not* recommded best-practice. The OnLoad / OnShown
methods (and matching events) are far more reliable.

Marc

 

[Moty Michaely]

public class FormEx : Form
{
public new void Show()
{
//do your pre-stuff
base.Show();
//do your post-stuff
}

public new void Show(IWin32Window owner)
{
//do your pre-stuff
base.Show(owner);
//do your post-stuff
}
}

Xavier PACOTTE je napisao/la:

Hi,

I wouldn't suggest using this solution since it's not polymorphic.

The following code will not do what you need:

Form frm = new Form1();
frm.Show();

Only if frm is of type Form1, the Form1 implementation of Show() will
be called.

Moty