Joe,

Why do you set the Form.Visible to .f. in the designer? The form doesn't become visible until the Show() (last night I was off by one method I said Activate()) is triggered by the internal code. The sequence is LISAG - Load, Init, Show, Activate, GotFocus.

I created a VFP baseclass form with this code in the LISAG methods:

debugout program()
this.Delay()

The Delay() method was

for i = 1 to 1000000
   j = i * 3
endfor

I opened the output window and the form does not become visible until the Show() occurs. You have to have some other code (somewhere in your class hierarchy maybe) causing your early visibility of the form.

>Thanks for the reply. That's what I thought, too, so I created an absolutely plain form, set the visible property to .F. and set step on in the first line of init(). thisform.visible is .F. in the debugger -- BUT the form is literally visible. I can see it.
>
>Moreover, I put a checkbox on the form with the controlsource set to thisform.visible. The checkbox is initally unchecked (this is after thisform.activate() executes), but I can see the form. If I click the checkbox once, it is now checked. Twice, and the form disappears.
>
>Clearly, this is not what I would expect. I'm sure I have some setting or other affecting this behaviour, but I have no clue what.