Use variable value as bytes array in Python

Question

I want to implement socket client in Python. The server expects the first 8 bytes contain total transmission size in bytes. In C client I did it like this:

uint64_t total_size = zsize + sizeof ( uint64_t );
uint8_t* xmlrpc_call = malloc ( total_size );
memcpy ( xmlrpc_call, &total_size, sizeof ( uint64_t ) );
memcpy ( xmlrpc_call + sizeof ( uint64_t ), zbuf, zsize );

Where zsize and zbuff are size and data I want to transmit. In python I create byte array like this:

cmd="<xml>do_reboot</xml>"
result = deflate (bytes(cmd,"iso-8859-1"))
size = len(result)+8

What is the best to fill the header in Python? Without separating value to 8 bytes and copy it in loop

Are you performing normal XML-RPC? Is there any reason why you can't use xmlrpclib? — Ignacio Vazquez-Abrams
– Ignacio Vazquez-Abrams, Commented Oct 29, 2014 at 11:42
It is custom XMLRPC, encrypted and zlibbed. I should support its protocol as is — pugnator
– pugnator, Commented Oct 29, 2014 at 12:09

Community · Accepted Answer · 2017-05-23 12:05:38Z

1

You could use the struct module, which will pack your data into binary data in the format you want

import struct
# ...your code for deflating and processing data here...

result_size = len(result)
# `@` means use native size, `I` means unsigned int, `s` means char[].
# the encoding for `bytes()` should be changed to whatever you need
to_send = struct.pack("@I{0}s".format(result_size), result_size, bytes(result, "utf-8"))

2 Comments

pugnator Over a year ago

Strange, but the header is always 0

XrXr Over a year ago

The header is there, you might be looking at the padding, which is 0. when result = "good day", to_send = b'\x08\x00\x00\x00good day'. You can also check using to_send[0]. If you want the padding to come before the number, you can use > instead of @

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