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pack a three byte int

P: n/a
Can Python not express the idea of a three-byte int?

For instance, in the working example below, can we somehow collapse the
three calls of struct.pack into one?
>>import struct

skip = 0x123456 ; count = 0x80

cdb = ''
cdb += struct.pack('>B', 0x08)
cdb += struct.pack('>I', skip)[-3:]
cdb += struct.pack('>BB', count, 0)

print ' '.join(['%02X' % ord(xx) for xx in cdb])
08 12 34 56 80 00
>>>
I ask because I'm trying to refactor working code that is concise:
>>cdb0 = '\x08' '\x12\x34\x56' '\x80' '\0'
print ' '.join(['%02X' % ord(xx) for xx in cdb0])
08 12 34 56 80 00
>>>
Thanks in advance, Pat LaVarre

Nov 8 '06 #1
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18 Replies


P: n/a
p.*******@ieee.org wrote:
Can Python not express the idea of a three-byte int?
It is a bit hard to determine what that (rhetorical?) question means.
Possible answers:
1. Not as concisely as a one-byte struct code -- as you presumably have
already determined by reading the manual ...
2. No, but when 24-bit machines become as popular as they were in the
1960s, feel free to submit an enhancement request :-)
>
For instance, in the working example below, can we somehow collapse the
three calls of struct.pack into one?
>import struct

skip = 0x123456 ; count = 0x80

cdb = ''
cdb += struct.pack('>B', 0x08)
cdb += struct.pack('>I', skip)[-3:]
cdb += struct.pack('>BB', count, 0)

print ' '.join(['%02X' % ord(xx) for xx in cdb])
08 12 34 56 80 00
>>
You could try throwing the superfluous bits away before packing instead
of after:

| >>from struct import pack
| >>skip = 0x123456; count = 0x80
| >>hi, lo = divmod(skip, 0x10000)
| >>cdb = pack(">BBHBB", 0x08, hi, lo, count, 0)
| >>' '.join(["%02X" % ord(x) for x in cdb])
| '08 12 34 56 80 00'

but why do you want to do that to concise working code???

Nov 8 '06 #2

P: n/a
In article <11**********************@k70g2000cwa.googlegroups .com>,
p.*******@ieee.org wrote:
Can Python not express the idea of a three-byte int?

For instance, in the working example below, can we somehow collapse the
three calls of struct.pack into one?
>import struct

skip = 0x123456 ; count = 0x80

cdb = ''
cdb += struct.pack('>B', 0x08)
cdb += struct.pack('>I', skip)[-3:]
cdb += struct.pack('>BB', count, 0)
Why not something like this:

skip += struct.pack(">L", skip)[1:]

Dave
Nov 9 '06 #3

P: n/a
Sorry, that should have been:

cdb += struct.pack(">L", skip)[1:]

Dave
Nov 9 '06 #4

P: n/a

Dave Opstad wrote:
Sorry, that should have been:

cdb += struct.pack(">L", skip)[1:]
">L" and ">I" produce exactly the same 4-byte result. The change from
[-3:] to [1:] is a minor cosmetic improvement, but obscures the
underlying ... a bit like putting mascara on a pig. I got the
impression that the OP was interested in more radical improvement.

Cheers,
John

Nov 9 '06 #5

P: n/a
Not as concisely as a one-byte struct code

Help, what do you mean?
you presumably... read... the manual ...
Did I reread the wrong parts? I see I could define a ctypes.Structure
since 2.5, but that would be neither concise, nor since 2.3.
when 24-bit machines become ... popular
Indeed the struct's defined recently, ~1980, were contorted to make
them easy to say in C, which makes them easy to say in Python, e.g.:

X28Read10 = 0x28
cdb = struct.pack('>BBIBHB', X28Read10, 0, skip, 0, count, 0)

But when talking the 1960's lingo I find I am actually resorting to
horrors like:

X12Inquiry = 0x12
xxs = [0] * 6
xxs[0] = X12Inquiry
xxs[4] = allocationLength
rq = ''.join([chr(xx) for xx in xxs])

Surely this is wrong? A failure on my part to think in Python?

Nov 9 '06 #6

P: n/a
cdb0 = '\x08' '\x01\x23\x45' '\x80' '\0'

cdb = ''
cdb += struct.pack('>B', 0x08)
cdb += struct.pack('>I', skip)[-3:]
cdb += struct.pack('>BB', count, 0)

The change from [-3:] to [1:] is a minor cosmetic improvement,
Ouch, [1:] works while sizeof I is 4, yes, but that's not what I meant,
more generally.

Something else I tried that doesn't work is:

skip = 0x12345 ; count = 0x80
struct.pack('>8b3b21b8b8b', 0x08, 0, skip, count, 0)

That doesn't work, because in Python struct b means signed char, not
bit; and a struct repeat count adds fields, rather than making a field
wider.

But do you see what I mean? The fields of this struct have 8, 3, 21,
8, and 8 bits. The "skip" field has 21 bits, the "count" field has 8
bits, I'd like to vary both of those.
why do you want to do that to concise working code???
cdb0 = '\x08' '\x01\x23\x45' '\x80' '\0'
works, but it's not parameterised. Writing out the hex literal always
packs the same x12345 and x80 values into those 21 and 8 bit fields.

I see I can concisely print and eval the big-endian hex:

X08Read6 = 0x08
skip = 0x12345 ; count = 0x80
hex = '%02X' % X08Read6 + ('%06X' % skip) + ('%02X' % count) + '00'
''.join([chr(int(hex[ix:ix+2],0x10)) for ix in range(0,len(hex),2)])

But that's ugly too. Maybe least ugly so far, if I bury the
join-chr-int-for-range-len-2 in a def.

Is there no less ugly way to say pack bits, rather than pack bytes, in
Python?

Nov 9 '06 #7

P: n/a
p.*******@ieee.org wrote:
Not as concisely as a one-byte struct code

Help, what do you mean?
Help, what did you mean by the question?

"struct" == "Python struct module"

Struct module has (concise) codes B, H, I, Q for unsigned integers of
lengths 1, 2, 4, 8, but does *not* have a code for 3-byte integers.
>
you presumably... read... the manual ...

Did I reread the wrong parts? I see I could define a ctypes.Structure
since 2.5, but that would be neither concise, nor since 2.3.
Looks like you ignored the first word in the sentence ("Not").
>
when 24-bit machines become ... popular

Indeed the struct's defined recently, ~1980, were contorted to make
them easy to say in C, which makes them easy to say in Python, e.g.:

X28Read10 = 0x28
cdb = struct.pack('>BBIBHB', X28Read10, 0, skip, 0, count, 0)

But when talking the 1960's lingo I find I am actually resorting to
horrors like:

X12Inquiry = 0x12
xxs = [0] * 6
xxs[0] = X12Inquiry
xxs[4] = allocationLength
rq = ''.join([chr(xx) for xx in xxs])

Surely this is wrong? A failure on my part to think in Python?
It looks wrong (and a few other adjectives), irrespective of what
problem it is trying to solve. Looks like little-endian 4-byte integer
followed by 2-byte integer ... what's wrong with struct.pack("<IH",
X12Inquiry, allocationLength) ????

Your original question asked about bigendian 3-byte integers; have you
read the suggested solution that I posted? Does it do what you asked
(one pack call instead of three)????

Nov 9 '06 #8

P: n/a
Help, what did you mean by the question?

How does Python express the idea:

i) Produce the six bytes '\x08' '\x01\x23\x45' '\x80' '\0' at run-time
when given the tuple (0x08, 0x12345, 0x80, 0).

ii) Produce the six bytes '\x12' '\0\0\0' '\x24' '\0' when given the
tuple (0x12, 0, 0x24, 0).

iii) And so on.

So far, everything I write is ugly. Help?
Looks like you ignored ...
I guess you're asking me to leave the mystery of my question alone long
enough to show more plainly that indeed I am trying to make sense of
every word of every answer.

I guess I should do that in separate replies, cc'ed back into this same
thread. Please stay tuned.

Thanks in advance, Pat LaVarre

Nov 10 '06 #9

P: n/a
"struct" == "Python struct module"
>
Struct module has (concise) codes B, H, I, Q for unsigned integers of
lengths 1, 2, 4, 8, but does *not* have a code for 3-byte integers.
I thought that's what the manual meant, but I was unsure, thank you.
1. Not as concisely as a one-byte struct code

Looks like you ignored the first word in the sentence ("Not").
I agree I have no confident idea of what your English meant.

I guess you're hinting at the solution you think I should find obvious,
without volunteering what that is.

Yes? If so, then:

I guess for you "a one-byte struct code" is a 'B' provided as a "format
character" of the fmt parameter of the struct.pack function.

Yes? if so, then:

You recommend shattering the three byte int:

skip = 0x012345 ; count = 0x80
struct.pack('>6B', 0x08, skip >0x10, skip >8, skip, count, 0)

Except you know that chokes over:

DeprecationWarning: 'B' format requires 0 <= number <= 255

So actually you recommend:

def lossypack(fmt, *args):
return struct.pack(fmt, *[(arg & 0xFF) for arg in args])
skip = 0x012345 ; count = 0x80
lossypack('>6B', 0x08, skip >0x10, skip >8, skip, count, 0)

Yes?
I guess you're asking me ...
to show more plainly that indeed I am trying
to make sense of every word of every answer
Am I helping?

Nov 10 '06 #10

P: n/a
when talking the 1960's lingo
...
X12Inquiry = 0x12
xxs = [0] * 6
xxs[0] = X12Inquiry
xxs[4] = allocationLength
rq = ''.join([chr(xx) for xx in xxs])

It looks wrong (and a few other adjectives),
Ah, we agree, thank you for saying.
Looks like little-endian 4-byte integer
followed by 2-byte integer ... what's wrong with struct.pack("<IH",
X12Inquiry, allocationLength) ????
Pack '<IH' doesn't match how the code that I'm refactoring thinks about
these things.

The people who wrote this stuff forty years ago were thinking of bit
fields - here bit lengths of 8 then 3 then 21 then 8 then 8 bits -
cheating only when the bit boundaries happened to hit byte boundaries.

Yes, as you describe in this example, I could cheat when the boundaries
happen to hit H or I boundaries as well, but then I'm still left coping
with the cases where the fields split on byte boundaries that are not H
or I boundaries, such as the example:
skip = 0x123456; count = 0x80
hi, lo = divmod(skip, 0x10000)

Does it do what you asked (one pack call instead of three)????
One pack call, not three, yes.

Shatters the 3 byte int into 1 and 2 bytes by divmod of (0xFFFF + 1),
yes.
I guess you're asking me ...
to show more plainly that indeed I am trying
to make sense of every word of every answer
Am I helping?

Nov 10 '06 #11

P: n/a
Speaking as the OP, perhaps I should mention:
[-3:] to [1:] is a minor cosmetic improvement
To my eye, that's Not an improvement.

'\x08' '\x01\x23\x45' '\x80' '\0' is the correct pack of (0x08,
0x12345, 0x80, 0) because '\x01\x23\x45' are the significant low three
bytes of a big-endian x12345, thus [-3:].

The [1:] fact that we can keep the 3 significant bytes by tossing
exactly 1 byte away after rounding the bit length of that digital
number up to the nearest power of two which happens to be 4 = 3 + 1 is
merely incidental - not of central significance.

Nov 10 '06 #12

P: n/a

p.*******@ieee.org wrote:
"struct" == "Python struct module"

Struct module has (concise) codes B, H, I, Q for unsigned integers of
lengths 1, 2, 4, 8, but does *not* have a code for 3-byte integers.

I thought that's what the manual meant, but I was unsure, thank you.
If it doesn't have a code for 3-byte integers in the table of codes, it
doesn't have one. What's to be unsure about??
>
1. Not as concisely as a one-byte struct code
Looks like you ignored the first word in the sentence ("Not").

I agree I have no confident idea of what your English meant.
"Not" is rather unambiguous.
>
I guess you're hinting at the solution you think I should find obvious,
without volunteering what that is.
I did volunteer what it is -- try reading the message again. You may
need to use the PageDown key :-)
>
Yes? If so, then:
No, not at all. Stop guessing. I ask again: have you read the solution
that I gave???
Here it is again:

"""
You could try throwing the superfluous bits away before packing instead
of after:

| >>from struct import pack
| >>skip = 0x123456; count = 0x80
| >>hi, lo = divmod(skip, 0x10000)
| >>cdb = pack(">BBHBB", 0x08, hi, lo, count, 0)
| >>' '.join(["%02X" % ord(x) for x in cdb])
| '08 12 34 56 80 00'

>
I guess for you "a one-byte struct code" is a 'B' provided as a "format
character" of the fmt parameter of the struct.pack function.

Yes?
Yes -- but I thought we were already over that.

if so, then:
This does not follow.
>
You recommend shattering the three byte int:
Yes, but not like that. See above.
>
skip = 0x012345 ; count = 0x80
struct.pack('>6B', 0x08, skip >0x10, skip >8, skip, count, 0)

Except you know that chokes over:

DeprecationWarning: 'B' format requires 0 <= number <= 255
That is not "choking" -- that is barfing; it is telling you that you
have done something silly.
>
So actually you recommend:

def lossypack(fmt, *args):
return struct.pack(fmt, *[(arg & 0xFF) for arg in args])
skip = 0x012345 ; count = 0x80
lossypack('>6B', 0x08, skip >0x10, skip >8, skip, count, 0)

Yes?
No, never, not in a pink fit.
>
I guess you're asking me ...
to show more plainly that indeed I am trying
to make sense of every word of every answer
You guess wrongly.
>
Am I helping?
No.

Nov 10 '06 #13

P: n/a
Perhaps Python can't concisely say three-byte int ...

But Python can say six-nybble hex:
>>import binascii
cdb = binascii.unhexlify('%02X%06X%02X%02X' % (0x08, 0x12345, 0x80, 0))
binascii.hexlify(cdb)
'080123458000'
>>>
Thanks again for patiently helping me find this. A shortcut is:

http://docs.python.org/lib/genindex.html
search: hex

Nov 10 '06 #14

P: n/a

p.*******@ieee.org wrote:

>
Pack '<IH' doesn't match how the code that I'm refactoring thinks about
these things.

The people who wrote this stuff forty years ago were thinking of bit
fields - here bit lengths of 8 then 3 then 21 then 8 then 8 bits -
cheating only when the bit boundaries happened to hit byte boundaries.

Yes, as you describe in this example, I could cheat when the boundaries
happen to hit H or I boundaries as well, but then I'm still left coping
with the cases where the fields split on byte boundaries that are not H
or I boundaries, such as the example:

Am I helping?
Yes, you have *finally* said unambiguously what your problem really is
-- field lengths not a multiple of 8 bits. I suggest that you start a
new thread, write it out logically and ask for assistance. You should
get some sensible answers. I will apologise in advance for not
participating; I'm exhausted.

Cheers,
John

Nov 10 '06 #15

P: n/a

p.*******@ieee.org wrote:
Speaking as the OP, perhaps I should mention:
[-3:] to [1:] is a minor cosmetic improvement

To my eye, that's Not an improvement.

'\x08' '\x01\x23\x45' '\x80' '\0' is the correct pack of (0x08,
0x12345, 0x80, 0) because '\x01\x23\x45' are the significant low three
bytes of a big-endian x12345, thus [-3:].

The [1:] fact that we can keep the 3 significant bytes by tossing
exactly 1 byte away after rounding the bit length of that digital
number up to the nearest power of two which happens to be 4 = 3 + 1 is
merely incidental - not of central significance.
I said *cosmetic* improvement and also said "obscures the underlying"
need-to-know that 4 - 3 == 1 (which I didn't contemplate needing 2
paragraphs of laborious explanation).

Nov 10 '06 #16

P: n/a
At Thursday 9/11/2006 22:24, p.*******@ieee.org wrote:
>Perhaps Python can't concisely say three-byte int ...

But Python can say six-nybble hex:
>import binascii
cdb = binascii.unhexlify('%02X%06X%02X%02X' % (0x08, 0x12345, 0x80, 0))
binascii.hexlify(cdb)
'080123458000'
The only problem I can see is that this code is endianness-dependent;
the suggested versions using pack(">...") not. But this may not be of
concern to you.
--
Gabriel Genellina
Softlab SRL

__________________________________________________
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Nov 10 '06 #17

P: n/a
... Python can say six-nybble hex:
>>import binascii
>>cdb = binascii.unhexlify('%02X%06X%02X%02X' % (0x08, 0x12345, 0x80, 0))
>>binascii.hexlify(cdb)
'080123458000'

The only problem I can see is that this code is endianness-dependent;
the suggested versions using pack(">...") not. But this may not be of
concern to you.
Thanks for cautioning us. I suspect we agree:

i) pack('>...') can't say three byte int.
ii) binascii.hexlify evals bytes in the order printed.
iii) %X prints the bytes of an int in big-endian order.
iv) struct.unpack '>' of struct.pack '<' flips the bytes of an int
v) struct.unpack '<' of struct.pack '>' flips the bytes of an int
vi) [::-1] flips a string of bytes.

In practice, all my lil-endian structs live by the C/Python-struct-pack
law requiring the byte size of a field to be a power of two, so I can
use Python-struct-pack to express them concisely. Only my big-endian
structs are old enough to violate that recently (since ~1972)
popularised convention, so only those do I construct with
binascii.unhexlify.

Often I wrap a big-endian struct in a lil-endian struct, but I'm ok
calling hexlify to make the big-endian struct and then catenating it
into the lil-endian struct, e.g., in the following cdb is big-endian
inside cbwBytes:

cbwBytes = struct.pack('<IIIBBB',
cbw.dSignature, cbw.dTag, cbw.dDataTransferLength,
cbw.bmFlags, cbw.bLun, cbw.bCbLength,
) + cdb[0:cdbLength] + ('\0' * (0x10 - cdbLength))

Nov 10 '06 #18

P: n/a
At Friday 10/11/2006 00:08, p.*******@ieee.org wrote:
>import binascii
>cdb = binascii.unhexlify('%02X%06X%02X%02X' % (0x08,
0x12345, 0x80, 0))
>binascii.hexlify(cdb)
>'080123458000'
The only problem I can see is that this code is endianness-dependent;
the suggested versions using pack(">...") not. But this may not be of
concern to you.

Thanks for cautioning us. I suspect we agree:

i) pack('>...') can't say three byte int.
ii) binascii.hexlify evals bytes in the order printed.
iii) %X prints the bytes of an int in big-endian order.
iv) struct.unpack '>' of struct.pack '<' flips the bytes of an int
v) struct.unpack '<' of struct.pack '>' flips the bytes of an int
vi) [::-1] flips a string of bytes.
Yes to all.
>In practice, all my lil-endian structs live by the C/Python-struct-pack
law requiring the byte size of a field to be a power of two, so I can
use Python-struct-pack to express them concisely. Only my big-endian
structs are old enough to violate that recently (since ~1972)
popularised convention, so only those do I construct with
binascii.unhexlify.
So you would have no problems. I stand corrected: the code above will
always generate big-endian numbers.
--
Gabriel Genellina
Softlab SRL

__________________________________________________
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Nov 10 '06 #19

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