"Evan" <xdi...@gmail.com> wrote in message
news:79582a34-5d0b-49b2-8c1e-4139324ff...@b38g2000prf.googlegroups.com...
Hello ~
I'm new with python, what my problem is, I have a binary file, I want
to read first 2 bytes and convert it to host byte order, then write it
to another file.
There is a piece of information missing here. What is the byte order of the
original binary file?
I try to use 'socket' and 'struct', but somehow i can not get it
working fine:
for example, totally I'm not sure if my steps is correct or not:
++++++++++++++++++++++++++++++++++++++++++++++++
import socket
f=open('a.bin','rb')
f.read(2)
'\x04\x00'
This is either a little-endian 4, or a big-endian 1024 (0x400).
f.seek(0)
st=f.read(2)
e=open('test.bin','w+b')
e.write(socket.ntohs(struct.unpack('H',st[:2])[0]))
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: argument 1 must be string or read-only buffer, not int
+++++++++++++++++++++++++++++++++++++++++++++++++
It failed due to the parameter is 'int', not 'str' in write function.
but how can i do that?
socket.ntohs returns an integer. write takes a string. ntohs assumes the
original value was big-endian (network) order.
If the original binary file is little-endian, this works:
import struct
f=open('a.bin','rb')
data = struct.unpack('<H',f.read(2))[0]
f.close()
e=open('test.bin','wb')
e.write(struct.pack('H',data)) # default is host-order, could be big or
little.
e.close()
If the original binary file is big-endian, change the 3rd line:
data = struct.unpack('>H',f.read(2))[0]
-Mark
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