Re: struct size confusion:
Michael Yanowitz wrote: 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? From TFM: Native size and alignment are determined using the C compiler's sizeof expression. This is always combined with native byte order. Standard size and alignment are as follows: no alignment is required for any type (so you have to use pad bytes); short is 2 bytes; int and long are 4 bytes; long long (__int64 on Windows) is 8 bytes; float and double are 32-bit and 64-bit IEEE floating point numbers, respectively See this how to achieve the desired results (on my system at least): print struct.calcsize(I) 4 print struct.calcsize(H) 2 print struct.calcsize(HI) 8 print struct.calcsize(=HI) 6 print struct.calcsize(=IH) 6 Regards, Diez -- http://mail.python.org/mailman/listinfo/python-list
Re: struct size confusion:
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 = HHHI
which is 34 bytes
Are you sure? Not for me:
In [9]: struct.calcsize('HHHI')
Out[9]: 36
Kent
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Re: struct size confusion:
Michael Yanowitz wrote: 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? because modern platforms tend to use an alignment equal to the size of the item; 2-byte objects are stored at even addresses, 4-byte objects are stored at addresses that are multiples of four, etc. in other words, HI is stored as 2 bytes H data plus 2 bytes padding plus four bytes I data, while IH is four bytes I data, no padding, and 2 bytes H data. I tried specifying big endian and little endian and they both have the same results. are you sure? (see below) I suspect, there is some kind of padding involved, but it does not seem to be done consistently or in a recognizable method. the alignment options are described in the library reference: http://docs.python.org/lib/module-struct.html default is native byte order, native padding: struct.calcsize(IH) 6 struct.calcsize(HI) 8 to specify other byte orders, use a prefix character. this also disables padding. e.g. struct.calcsize(!IH) 6 struct.calcsize(!HI) 6 /F -- http://mail.python.org/mailman/listinfo/python-list
RE: struct size confusion:
Thanks for your and everyone else's feedback. I got it to work now by prefixing the PACK_FORMAT with !. I previously thought I could only use the !' with the unpack. I still don't fully understand the byte allignment stuff (I am sure I will get it eventually), but I am content that it is working now. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of Fredrik Lundh Sent: Wednesday, March 22, 2006 9:28 AM To: python-list@python.org Subject: Re: struct size confusion: Michael Yanowitz wrote: 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? because modern platforms tend to use an alignment equal to the size of the item; 2-byte objects are stored at even addresses, 4-byte objects are stored at addresses that are multiples of four, etc. in other words, HI is stored as 2 bytes H data plus 2 bytes padding plus four bytes I data, while IH is four bytes I data, no padding, and 2 bytes H data. I tried specifying big endian and little endian and they both have the same results. are you sure? (see below) I suspect, there is some kind of padding involved, but it does not seem to be done consistently or in a recognizable method. the alignment options are described in the library reference: http://docs.python.org/lib/module-struct.html default is native byte order, native padding: struct.calcsize(IH) 6 struct.calcsize(HI) 8 to specify other byte orders, use a prefix character. this also disables padding. e.g. struct.calcsize(!IH) 6 struct.calcsize(!HI) 6 /F -- http://mail.python.org/mailman/listinfo/python-list -- http://mail.python.org/mailman/listinfo/python-list
