> I gather that you mean '3 lots of 4 _bits_' :-)

Oops

> You will have heard by now that the above routine was a 2-digit packed
BCD-to-Binary routine ...

Oops again

> If you are only counting (not converting the output of a calculation) you
> could opt for counting BCD-wise. said otherwise : increment the BCD-
> counter and BCD-adjust the result.

I did ponder this one; I could do it either way, but I thought that the binary
to BCD conversion would be a solution that I'd use again.

> I've got a Binary-to-BCD routine using a long-tail division (homebrew) for
> two bytes (resulting in 5 digits uncompressed BCD). If you are interrested

I sure am. Thoroughly commented, I trust? I'm still trying to puzzle out how
it works. Horowitz and Hill just say 'It's complicated' which is not helpful. I'
ve
even drawn up a K-map and stared at it for a while. I stopped when I got a
headache.

I'd be most grateful for sight of your routine.

Kind regards

John M