How to pack a UUID into a struct in Python?

问题

I have a UUID that I was thinking of packing into a struct using UUID.int, which turns it into a 128-bit integer. But none of the struct format characters are large enough to store it, how to go about doing this?

Sample code:

s = struct.Struct('L')
unique_id = uuid.uuid4()    
tuple = (unique_id.int)
packed = s.pack(*tuple)

The problem is, struct format 'L' is only 4 bytes...I need to store 16. Storing it as a 32-char string is a bit much.

回答1:

It is a 128-bit integer, what would you expect it to be turned into? You can split it into several components — e.g. two 64-bit integers:

max_int64 = 0xFFFFFFFFFFFFFFFF
packed    = struct.pack('>QQ', (u.int >> 64) & max_int64, u.int & max_int64)
# unpack
a, b     = struct.unpack('>QQ', packed)
unpacked = (a << 64) | b

assert u.int == unpacked

回答2:

As you're using uuid module, you can simply use bytes member, which holds UUID as a 16-byte string (containing the six integer fields in big-endian byte order):

u = uuid.uuid4()
packed = u.bytes # packed is a string of size 16
assert u == uuid.UUID(bytes=packed)

来源：https://stackoverflow.com/questions/6877096/how-to-pack-a-uuid-into-a-struct-in-python