Using big endian in struct pack Python with a Pointer

Question

I'm using the struct pack in Python 3.7.
I have this header containing 6 uint32 and one void * argument.

I want to put it in binary format. Unfortunately the interface I'm working on only supports big endian.So I actually need to transform what I send. But since P format can't have its endianess modified I always get struct error. So how can I rewrite this to avoid having this error.

PS: I cannot remove the void * it an mandatory field of my header

u1 = 0x4569 
u2 = 0x1236 
u3 = 0x4411 
u4 = 0x1236 
u5 = 0x9696 
point = 0 #(For now) 
data = 0x26358974 
buffer = struct.pack('5IPI',u1,u2,u3,u4,u5,point,data)

Actual output

buffer = b'\x69\x45\x00\x00\x36\x12\x00\x00\x11\x44\\x00\x00\x36\x12\x00\x00\x96\x96\\x00\x00\x00\x00\x00\x00\x74\x89\x35\26'

WHILE what I want would be:

buffer = b'\x00\x00\45\x69\x00\x00\x12\36\x00\x00\x44\x11\x00\x00\x12\x36\\x00\x00\x96\x96\x00\x00\x00\x00\x26\x35\x89\x74'

Please update the question with example input and required output. — quamrana
– quamrana, Commented May 31, 2019 at 8:33
The documentation says that P is only available for native byte ordering. — quamrana
– quamrana, Commented May 31, 2019 at 8:42
u1 = 0x4569 u2 = 0x1236 u3 = 0x4411 u4 = 0x1236 u5 = 0x9696 point = 0 (For now) data = 0x26358974 In the current state doing: buffer = struct.pack('5IPI',u1,u2,u3,u4,u5,point,data) give me : buffer = b'\x69\x45\x00\x00\x36\x12\x00\x00\x11\x44\\x00\x00\x36\x12\x00\x00\x96\x96\\x00\x00\x00\x00\x00\x00\x74\x89\x35\26' WHILE what I want would be: buffer = b'\x00\x00\45\x69\x00\x00\x12\36\x00\x00\x44\x11\x00\x00\x12\x36\\x00\x00\x96\x96\x00\x00\x00\x00\x26\x35\x89\x74' — JenG18
– JenG18, Commented May 31, 2019 at 9:10

quamrana · Accepted Answer · 2019-05-31 09:15:38Z

1

You may need to do this in two stages:

u1=1
u2=2
u3=3
u4=4
u5=5
point=6
data=7

# construct two separate buffers
b1 = struct.pack('>5I', u1, u2, u3, u4, u5)
b2 = struct.pack('P', point)

Then convert the pointer back to a regular uint, and then back to a pointer:

p2 = struct.unpack('I', b2)
b2a = struct.pack('>I', p2[0])
b1 = b1 + b2a
print(b1)

Output:

b'\x00\x00\x00\x01\x00\x00\x00\x02\x00\x00\x00\x03\x00\x00\x00\x04\x00\x00\x00\x05\x00\x00\x00\x06'

answered May 31, 2019 at 9:15

quamrana

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8 Comments

JenG18 Over a year ago

Thanks for your help. I get this error now : p2 = struct.unpack('I', b2) struct.error: unpack requires a buffer of 4 bytes

quamrana Over a year ago

My b2 is b'\x06\x00\x00\x00' which is 4 bytes (little endian of course). What do you have?

JenG18 Over a year ago

my b2 is b'\x00\x00\x00\x00\x00\x00\x00\x00'. In the python struct officiel page there is no standard size so maybe i need to force the size myself?

quamrana Over a year ago

Sounds like you are on a 64 bit system. What does this pointer mean once it is packed?

JenG18 Over a year ago

It's a reserved field in order to prevent user to enter unwanted line command in my system.

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