struct.calcsize problem

[Bengt Richter]

On 7 Nov 2005 15:27:06 -0800, "Chandu" <chandu at trillian.us> wrote:

>In using the following struct format I get the size as 593. The same C
>struct is 590 if packed on byte boundary and 596 when using pragma
>pack(4). I am using pack(4) and added 3 spares at the end to get by.
>   hdrFormat = '16s 32s 32s B 8s H 8s H 4s H H 20s 64s 64s 64s 32s 32s
>64s L L B B B B B 64s 64s'
>
>Any ideas on what I am doing wrong?
>
Looks to me like you are getting default native byte order and _alignment_
and some pad bytes are getting added in. For native order with no padding,
try prefixing the format string with '='

 
 >>> hdrFormat
 '16s 32s 32s B 8s H 8s H 4s H H 20s 64s 64s 64s 32s 32s 64s L L B B B B B 64s 64s'

If you add up the individual sizes, padding doesn't happen, apparently:

 >>> map(struct.calcsize, hdrFormat.split())
 [16, 32, 32, 1, 8, 2, 8, 2, 4, 2, 2, 20, 64, 64, 64, 32, 32, 64, 4, 4, 1, 1, 1, 1, 1, 64, 64]
 >>> sum(map(struct.calcsize, hdrFormat.split()))
 590

But default:
 >>> struct.calcsize(hdrFormat)
 593
Apparently is native, with native alignment & I get the same as you:
 >>> struct.calcsize('@'+hdrFormat)
 593
Whereas native order standard (no pad) alignment is:
 >>> struct.calcsize('='+hdrFormat)
 590
Little endian, standard alignment:
 >>> struct.calcsize('<'+hdrFormat)
 590
Big endian, standard alignment
 >>> struct.calcsize('>'+hdrFormat)
 590
Network (big endian), standard alignment:
 >>> struct.calcsize('!'+hdrFormat)
 590

I guess if you want alignment for anything non-native, you have to specify pad bytes
where you need them (with x format character).

Regards,
Bengt Richter