python - Why does the output of struct.pack depend on the order of the items?

Question

I'm trying to pack an epoch with a string & I can't figure out why the number of bytes depends on the order I do the packing in. Unfortunately I'm not getting any replies on the relevant microcontroller forum https://forum.pycom.io/topic/6761/struct

If I convert a character to bytes I get 1 byte:

>>> bytes=struct.pack('s', 'F'); print(bytes, len(bytes))
b'F' 1

If I convert an epoch to bytes I get 4 bytes:

>>> bytes=struct.pack('I', 1611017052); print(bytes, len(bytes))
b'\+x06`' 4

How come when I do both together I get 8 instead of 4+1=5 ??

>>> bytes=struct.pack('sI', 'F', 1611017052); print(bytes, len(bytes))
b'Fx00x00x00\+x06`' 8

but this way I get the 5 I expect?

>>> bytes=struct.pack('Is', 1611017052, 'F'); print(bytes, len(bytes))
b'\+x06`F' 5

why is a different packing sequence giving different numbers of bytes?

question from:https://stackoverflow.com/questions/65836735/why-does-the-output-of-struct-pack-depend-on-the-order-of-the-items

Answer 1

It is a question of alignment. By default, struct packs according to standard C language conventions. This means that integers which are 4 bytes long are aligned on 4 byte boundaries. The start of the structure buffer is already aligned, so putting an integer first cause no padding. Putting a character first changes the offset to 1, so 3 bytes of padding are needed to get to a 4 byte boundary for a following integer.

This sort of padding is usually desirable, as some architectures may not allow the cpu to do an unaligned integer access, and others may do so much more slowly than an aligned access.

You can override the alignment by beginning the format with =:

bytes = struct.pack('=sI', b'F', 1611017052)