struct size confusion:

[Kent Johnson]

Michael Yanowitz wrote:
> Hello:
> 
>    I am relatively new to Python and this is my first post on
> this mailing list.
> 
>    I am confused as to why I am getting size differences in the following
> cases:
> 
> 
>>>>print struct.calcsize("I")
> 
> 4
> 
>>>>print struct.calcsize("H")
> 
> 2
> 
>>>>print struct.calcsize("HI")
> 
> 8
> 
>>>>print struct.calcsize("IH")
> 
> 6
> 
>    Why is it 8 bytes in the third case and why would it be only 6 bytes
> in the last case if it is 8 in the previous?

By default the struct module uses native byte-order and alignment which 
may insert padding. In your case, the integer is forced to start on a 
4-byte boundary so two pad bytes must be inserted between the short and 
the int. When the int is first no padding is needed - the short starts 
on a 2-byte boundary.

To eliminate the padding you should use any of the options that specify 
'standard' alignment instead of native:

In [2]: struct.calcsize('I')
Out[2]: 4

In [3]: struct.calcsize('H')
Out[3]: 2

In [4]: struct.calcsize('HI')
Out[4]: 8

In [5]: struct.calcsize('IH')
Out[5]: 6

In [6]: struct.calcsize('!HI')
Out[6]: 6

In [7]: struct.calcsize('>HI')
Out[7]: 6

In [8]: struct.calcsize('<HI')
Out[8]: 6
> 
>    I tried specifying big endian and little endian and they both have
> the same results.

Are you sure? They should use standard alignment as in the example above.
> 
>     I suspect, there is some kind of padding involved, but it does not
> seem to be done consistently or in a recognizable method.
> 
>    I will be reading shorts and longs sent from C into Python
> through a socket.
> 
>    Suppose I know I am getting 34 bytes, and the last 6 bytes are a 2-byte
> word followed by a 4-byte int, how can I be assured that it will be in that
> format?
> 
>    In a test, I am sending data in this format:
>   PACK_FORMAT = "HBBBBHBBBBBBBBBBBBBBBBBBBBHI"
> which is 34 bytes

Are you sure? Not for me:

In [9]: struct.calcsize('HBBBBHBBBBBBBBBBBBBBBBBBBBHI')
Out[9]: 36

Kent