问题
Please give me sample code to generate UUID of long type in java without using timestamp.
Thanks
回答1:
A real UUID is 128 bits. A long is 64 bits.
This is not just pedantry. UUID stands for Universal Unique IDentifier.
The "universal uniqueness" of the established UUID schemes are based on:
- encoding a MAC address and a timestamp,
- encoding a hash of a DNS name and a timestamp, or
- using a 122 bit random number ... which is large enough that the probability of a collision is very very small.
With 64 bits, there are simply not enough bits for "universal uniqueness". For instance, the birthday paradox means that if we had a number of computers generating random 64 bit numbers, the probability of a potentially detectable collision would be large enough to be of concern.
Now if you just want a UID (not a UUID), then any 64-bit sequence generator will do the job, provided that you take steps to guard against the sequence repeating. (If the sequence repeats, then the IDs are not unique in time; i.e. over time a given ID may denote different entities.)
回答2:
Have you looked at java.util.UUID?
回答3:
If you just want a simple unique long you can use AtomicLong.incrementAndGet(). This doesn't use a timestamp but does reset to 0 every time you start it and is not unique across JVMs.
What is the requirement not to use timestamps all about? UUID uses a timestamp. (amoungst other things)
来源：https://stackoverflow.com/questions/4529727/generate-uuid-of-long-type