How to unpack 6 bytes as a whole using struct in Python

Question

I have the following 8 bytes:

b'\x05\x00\x00\x00\x00\x00\x05\x00' 

I am trying to get two integers using struct.unpack: one for the first 2 bytes and one for the last 6. Getting the first two is easy with:

 struct.unpack("<H6B") 

But it returns

 (5, 0, 0, 0, 0, 5, 0) 

I want it to return the following:

 (5, 5) 

How to get integer value from last 6 bytes? I do not want each byte to be individual.

Answer 1

struct does not support non-two-two integers. This is a common occurrence. C does not support such integers on your platform (well, bit fields, but you cannot create an array of them).

 def unpack48(x): x1, x2, x3 = struct.unpack('<HHI', x) return x1, x2 | (x3 << 16) 

Answer 2

The standard structural module does not support all possible sizes, so you need to either combine several bits (see Dietrich's answer), or you can use external modules such as bitstring .

 >>> from bitstring import BitArray >>> b = BitArray(bytes=b'\x05\x00\x00\x00\x00\x00\x05\x00') >>> b.unpack('<H6B') [5, 0, 0, 0, 0, 5, 0] 

which is consistent with the standard struct.unpack . But now we can instead unpack the second element as a 6-byte (48-bit) unsigned unsigned integer:

 >>> b.unpack('<H, uintle:48') [5, 21474836480] 

which gives you the same result as Dietrich’s answer, and also shows that you have something wrong with your question! What do you need in this case:

 >>> b.unpack('uintle:48, <H') [5, 5] 

Note that you could also write <H as uintle:16 if you need a more consistent write.