03-07-2003, 06:08 PM
I probably should leave this one for Jarulf, but this is what I think would happen: Step 2 actually generates a random unsigned integer, divides it by [clvl/qlvl(book)], takes the remainder, and adds 1. With a negative divisor, the results are implementation defined, so I'm not absolutely positive what would happen. What I think would happen in this case is that the remainder will always be considered positive. So the end result probably is that the results using a -1 qlvl are identical to the results using a +1 qlvl.
There are still a few ways to crash Diablo, but I don't think this is one of them.
There are still a few ways to crash Diablo, but I don't think this is one of them.