python: unexpected behavior from struct pack()

Question

I'm having trouble using the struct.pack() for packing an integer.

With

struct.pack("BIB", 1, 0x1234, 0) 

I'm expecting

'\x01\x00\x00\x034\x12\x00'

but instead I got

'\x01\x00\x00\x004\x12\x00\x00\x00'

I'm probably missing something here. Please help.

Answer 1

'\x01\x00\x00\x004\x12\x00\x00\x00'
 ^ this '4' is not part of a hex escape

is actually the same as:

'\x01\x00\x00\x00\x34\x12\x00\x00\x00'

Because the ASCII code for "4" is 0x34.

Because you used the default (native) format, Python used native alignment for the data, so the second field was aligned to offset 4 and 3 zeroes were added before it.

To get a result more like what you wanted, use the format >BIB or <BIB (for big-endian or little-endian respectively) This gives you '\x01\x00\x00\x12\x34\x00' or '\x01\x34\x12\x00\x00\x00'. Neither of those are exactly what you specified, because the example you gave was not proper big-endian or little-endian representation of 0x1234.

See also: section Byte Order, Size, and Alignment in the documentation.

Answer 2

From the docs

Note By default, the result of packing a given C struct includes pad bytes in order to maintain proper alignment for the C types involved; similarly, alignment is taken into account when unpacking. This behavior is chosen so that the bytes of a packed struct correspond exactly to the layout in memory of the corresponding C struct. To handle platform-independent data formats or omit implicit pad bytes, use standard size and alignment instead of native size and alignment: see Byte Order, Size, and Alignment for details.

You can get your desired result by forcing the byte order. (chr(0x34) == '4')

>>> struct.pack(">BIB", 1, 0x1234, 0)
'\x01\x00\x00\x124\x00'