| msg110201 - (view) |
Author: hannes reuter (hannes.reuter) |
Date: 2010-07-13 15:00 |
| on http://docs.python.org/library/struct.html in section 7.3.2.2. Format Characters in the table one can find that L -> unsigned long -> integer-> should have a 4 byte dataspace. (same applies for l), which seems to be correct -> 32bit/8=4 On a Ubuntu 64 , Python 2.6, a) calcsize reports it uses 8 byte b) by writing data to a file and reading it back in I could confirm that it uses as well 8 bytes of binary data -> ofile2.write(struct.pack('LL',int(x[0]), int(x[1]))) Q: Is that behavior what we see a Docu error ? Q: Is that behavior what we see general struct problem ? add1: in # 2263 e.g. we see problems for L type data in relation to np ? add2: IMHO certainly not a padding problem like in #7355 for reproduction: from struct import * calcsize('H') 2 -> correct to docu calcsize('f') 4 -> correct to docu calcsize('L') 8 -> should be 4 calcsize('l') 8 -> should be 4 calcsize('x') 1 -> correct to docu So either docu is wrong, or struct has an error. |
|
|
| msg110223 - (view) |
Author: Mark Dickinson (mark.dickinson) *  |
Date: 2010-07-13 19:57 |
| Please read the three sentences directly preceding that table and tell me whether they clear this up for you. |
|
|
| msg110278 - (view) |
Author: hannes reuter (hannes.reuter) |
Date: 2010-07-14 13:16 |
| Dear Marc, Thanks for taking time to answer that question. I understand that this comes from the native formating i specified, >>> calcsize('L') 8 >>> calcsize('<L') 4 So if written with < or = it is 4 bytes, clear. However, as my system is a little endian one(e.g. sys.byteorder=little), whats the difference between native and little ? I understand that Alignment is not performed on the Native format, a) but why is only L/l shows a difference compared to the table(docu)/formated output in the in native format , while the rest agrees one to one ? b) Where could I look up/find such a native format table ? Probably you can answer that easily to me, I'm just really,really puzzled. Hannes for x in list: ... print x,calcsize(x),calcsize('<'+x) ... x 1 1 c 1 1 b 1 1 B 1 1 ? 1 1 h 2 2 H 2 2 i 4 4 I 4 4 l 8 4 L 8 4 q 8 8 Q 8 8 f 4 4 d 8 8 s 1 1 p 1 1 'little' On 7/13/10, Mark Dickinson <report@bugs.python.org> wrote: > > Mark Dickinson <dickinsm@gmail.com> added the comment: > > Please read the three sentences directly preceding that table and tell me > whether they clear this up for you. > > ---------- > assignee: theller -> mark.dickinson > nosy: +mark.dickinson > > _______________________________________ > Python tracker <report@bugs.python.org> > <http://bugs.python.org/issue9249> > _______________________________________ > |
|
|
| msg110312 - (view) |
Author: Mark Dickinson (mark.dickinson) * |
Date: 2010-07-14 18:59 |
| > However, as my system is a little endian one(e.g. > sys.byteorder=little), whats the difference between native and little Native mode uses: native size, native byteorder and alignment that matches your platform Little endian: standard size, little-endian, no alignment The *native* size means the size of the corresponding C type (e.g., as computed by sizeof) on your platform. So on a typical 64-bit Unix-alike platform, that's 8 for 'l' and 'L'; on 64-bit Windows and most 32-bit platforms, it's 4 for 'l' and 'L'. The *standard* size is as given by the table. It's the same on all platforms. It's true that on most common platforms the 'l' and 'L' codes are the only ones likely to differ. > b) Where could I look up/find such a native format table ? Why not just use struct.calcsize? This is all explained in the docs; I'm going to close this issue, since I don't think there's any discrepancy between the docs and the behaviour of the module. However, if you have ideas for specific improvements to the documentation, please do open another issue. |
|
|
| msg110367 - (view) |
Author: hannes reuter (hannes.reuter) |
Date: 2010-07-15 13:59 |
| Hi Mark, If you would add a footnote to the L/l formats table and mention what you wrote, it would make things clearer. Something along the lines like that I reformulated from your explanation: On most common platforms the 'l' and 'L' codes are the only ones likely to differ due to formating differences between Native and Little are. The Size for 'l' and 'L' on a typical 64-bit Unix-alike platform will be 8 for 'l' and 'L'; on 64-bit Windows and most 32-bit platforms, it's 4. This is due to the fact that in Native mode native size, native byteorder and alignment will be set, while with a Little endian formating standard size, little-endian, no alignment will be set. << Cheers Hannes On 7/14/10, Mark Dickinson <report@bugs.python.org> wrote: > > Mark Dickinson <dickinsm@gmail.com> added the comment: > >> However, as my system is a little endian one(e.g. >> sys.byteorder=little), whats the difference between native and little > > Native mode uses: native size, native byteorder and alignment that > matches your platform > Little endian: standard size, little-endian, no alignment > > The *native* size means the size of the corresponding C type (e.g., as > computed by sizeof) on your platform. So on a typical 64-bit Unix-alike > platform, that's 8 for 'l' and 'L'; on 64-bit Windows and most 32-bit > platforms, it's 4 for 'l' and 'L'. > > The *standard* size is as given by the table. It's the same on all > platforms. > > It's true that on most common platforms the 'l' and 'L' codes are the only > ones likely to differ. > >> b) Where could I look up/find such a native format table ? > Why not just use struct.calcsize? > > This is all explained in the docs; I'm going to close this issue, since I > don't think there's any discrepancy between the docs and the behaviour of > the module. > > However, if you have ideas for specific improvements to the documentation, > please do open another issue. > > ---------- > status: open -> closed > > _______________________________________ > Python tracker <report@bugs.python.org> > <http://bugs.python.org/issue9249> > _______________________________________ > |
|
|