A 32-bit long integer should be more than enough, allowing the counter to reach 2,147,483,647 before the database would wrap around. Assuming a 5-minute average song, you can only play 10,5120 songs in a year if the player plays nonstop. By my math, it would take 20,428 years to roll the counter over

Ah, Tony, you knew I couldn't leave this one alone, didn't you.

First, I think that a 32 bit integer would allow you 4,294,967,295 numbers (assuming we don't count zero as one of the set, that is...). At least, that's what 2^32 comes out to. 2,147,483,647 is 2^16. But maybe I don't understand integer mapping all that well, and a 32 bit integer really only gives you 2^16 numbers?

However, a much more serious problem is that you neglected to factor in the leap years. By my rough calculations, you would actually only get 20,415 years by setting your average year equal to 365.25 days. Of course, this does not take into account the leap year exceptions at 100 years, 400 years, and 1000 years, which would add about another half year to the total.

So, given the fact that we would get 14 fewer years than you calculated, do you think that a 32 bit integer would be enough? (That's 14 fewer because you rounded down from 20,428.8, should have rounded up to 20,429).

Of course, if the 32 bit integer really does give 4,294,967,295 numbers, then the whole question becomes moot, doesn't it?

tanstaafl.



"There Ain't No Such Thing As A Free Lunch"