[Solved] Access violation trying to create a timer

[Wilko500]

Following on from my last post https://forum.lazarus.freepascal.org/index.php/topic,71422.0.html the timer works as it should when started by a button on the main form, starts the timer and displays the date/time on main form

Code: Pascal  [Select][+][-]

1. procedure TfrmMain.Button1Click(Sender: TObject);
2. Var
3.     MyTest: TMyTimer;
4. begin
5.     MyTest.Create;
6.     MyTest.Start(1000, @DisplayTime);
7. end;  

But I want it to start automatically when program starts so I put a call at the end of form.Activate and then got an Access violation.  I had something like this before when I tried to do something before the form was fully loaded/ready so I recoded to call the same code with Application.QueueAsyncCall at the end of Form.Activate.  This calls the same code

Code: Pascal  [Select][+][-]

1. Procedure TfrmMain.StartTimers;
2. Var
3.     MyTest: TMyTimer;
4. begin
5.     MyTest.Create;  //<=  Access violation here
6.     MyTest.Start(1000, @DisplayTime);
7. End;{Procedure Timers}

which still gives an access violation

Code: Pascal  [Select][+][-]

1. exception class 'Process stopped with reason: EXC_BAD_ACCESS (code=EXC_I386_GPFLT)'.
2.  
3. In file 'timerunit.pas' at line 30:
4. FTimer := TTimer.Create(nil);

I am very puzzled as to why the same code works from a button but fails when run from Application.QueueAsyncCall.

[dseligo]

You don't create TMyTimer correctly.

Do it like this:

Code: Pascal  [Select][+][-]

1. Procedure TfrmMain.StartTimers;
2. Var
3.     MyTest: TMyTimer;
4. begin
5.     MyTest := TMyTimer.Create;  // <=  see this line
6.     MyTest.Start(1000, @DisplayTime);
7. End;{Procedure Timers}

[Wilko500]

Thank you @dseligo.  That's a duh-rr moment! And of course it's obvious when I look at it now.  Because the incorrect code, that I copied,  was working I was looking for a more complicated solution rather than the obvious. And it turns out that I can run the StartTimers procedure from form.Activate as I wanted to.