"You haven't declared your loop variable volatile. This means that the compiler can optimise it away completely if it sees fit. Even if it doesn't do so the time taken to execute the loop may well vary between compiler versions and optimisation levels. In a nutshell, about the only thing you can say for sure about it is that it is extremely unlikely to give the delay you expect."

I think you'll find that this is a routine that lots of people have used without complaint. I will, however, take on-board your suggestion that the loop variable might be better declared as volatile, despite your insistence that the routine is totally worthless. There is certainly no cost in doing so, and it might prove cheap insurance to make this cautionary change.

Regards,
Murray R. Van Luyn.