The problem is with the guarantees that the method does. This is a valid implementation for random number generator: http://xkcd.com/221/
A UUID isn't going to be duplicated, because it is unique to the machine and time it was generated, and there is care taken to ensure that even if you have two thread creating the a guid on the same micro second, you'll get two different guids. There is no such guarantees for random numbers. On Sat, Oct 4, 2008 at 3:06 AM, Paul Davis <[EMAIL PROTECTED]>wrote: > Ayende, > > Perhaps I'm missing something, but I don't see the problem. > > I suppose you could technically claim that we're not setting the > version bits properly but that seems rather not important given that > we have the revision protection. Also unless I'm mistaken this would > only be a problem if someone is using an external UUID generation > scheme that happens to be extremely not random. Which while possible, > would still send up red flags on constant collisions on document > creation. > > Paul > > On Fri, Oct 3, 2008 at 7:49 PM, Ayende Rahien <[EMAIL PROTECTED]> wrote: > > Looking at the method implementation, it looks like there might be a > problem > > here. > > > > new_uuid() -> > > list_to_binary(to_hex(crypto:rand_bytes(16))). > > > > In particular, we aren't actually guaranteed to have a unique value. > > You can read more about what needs to be done to get unique guids: here: > > http://blogs.msdn.com/oldnewthing/archive/2008/06/27/8659071.aspx > > >
