Int 5 (& "longer" code)
 

I havent tried this yet, but am thinking theoreticaly for now.

What happens if code that we set in int 5 takes more than one vertical blank to be finished?

It will be executed again upon exiting the irq service routine.

Meaning what? At second vbl it wont be called again cause its already being executed?

An IRQ of the same priority cannot interrupt a running IRQ, hence, the VBL can't interrupt the VBL still being executed. Since the interrupt source is cleared at the beginning of the interrupt handler and the new VBL sets it again, in the moment the interrupt routine quits (and the IRQ status flag is cleared), the next VBL will be triggered. This would also happen if the raster beam is now already drawing a new frame onto the monitor such that the VBL would be delayed. If the VBL routine runs for two frames or more, one VBL will be dropped.

Old games often had almost all game code in the VBL routine so that it was synced from the VBL. This is not advisable and a rather ugly hack. Of course, this must not be done at all if you have the OS running.

Aha, ok. Thanks for explanation.

For what reason is that an ugly hack?
It may be not advisable if you don't manage to run your engine in 50fps. But that's all.

Yes, you are right. If you are alone in the system, it is ok. I guess the fact that you are blocking interrupts of lower priority if you do calculations that need not be carried out with priority (which should be much of the game logic) doesn't really play a role in an Amiga game running as the only program.