On Tue, 30 Apr 2002, Drylar Levre wrote: > Hrm. I haven't contributed anything of value in a long time, so here... > (it wouldn't let me attach and send, so it's a kinda long one) > > A while back I was going to start a mud, but scrapped the whole lot soon > after > attempting a few major overhauls at once. I had put in some autosetting of > mobiles, which worked decently well. Since then I decided to tackle the job > from > the start again, this time I am working with more success. I decided to rip > the old autosetting out of the code I used to have and apply it to the new > mud, > and here is my product.
[SNIPPED] > /*calculate easy ac values*/ > ac[1] = 99 - (arg2 * 6.37) - ((arg2/10)^2); I have to admit, I've never seen a bitwise xor (^) used in quite this way. Could you explain why you're either adding or subtracting 2 to (arg2/10)? I don't quite understand it. > bonus = (arg2/59 + 1) * ((arg2 * 9) + (arg2/11)) +1; If I were to write code to do something similar to this, I think I would probably use floating point division to smooth out the curve instead of using a step function. Is there a reason you used integer division in all but a few places? Dennis
