Python Forum

as topics. i want to read a binary file. i need some some offset in it and control by big or little endian. but i have problem with it.

path_xd = 'xxxxx'
pattern = b'\x14\x02'
reg_xd=re.compile(pattern)
with open(path_xd, 'r+b') as p_xd:
    data=p_xd.read()
    for x in reg_xd.finditer(data):
        offset=x.start()
        p_xd.seek(offset)
        data_1402=p_xd.read(16)
        xd_name=int.from_bytes(data_1402[2:4], byteorder='little')
        xd_name=(str(xd_name).replace('0', '')).zfill(2)
        print(xd_name)
        print(data_1402)
        print(data_1402[5:9])

i get output like this

Output:
b'\x14\x02\x01\x00 \x00\x00\x00\x03\x18\x90BW#\x08\x00'

it's how it look like in termial

Output:
402 0100 2000 0000 0318 9042 5723  [..... ......BW#
00000040: 0800 5723 0800 

it seems missing some bytes in python output.

i want result like this format. for fullfile. seems i have misunderstand something here. need help with file hex output and big/litte endian please advice.thanks.

Output:
\x14\x02\x01\x00\x20\x00\x00\x00\x03\x18\x90\x42\x57\x23\x08\x00\x57\x23\x08\x00

up. anyone can help? thanks!

I don't understand when your output appears. The first one says you get "output like that", and the second one says it's how it looks in the terminal. I would expect your output to be in a terminal. How do you produce the first and how do you produce the second?

Do you have an example of the 'xxxxx' file that someone could execute your code against?

Looks like at least partly you have a byte string. When printed by default, if the byte maps to an ASCII character (like \x20), then it prints that character. \x20 is a space, so the 20s are removed and replaced with spaces in your printout.

If you just need the hex values, you could use (among other things) the hex method.

>>> b = b'\x14\x02\x01\x00\x20\x00\x00\x00\x03\x18\x90\x42\x57\x23\x08\x00\x57\x23\x08\x00'
>>> b
b'\x14\x02\x01\x00 \x00\x00\x00\x03\x18\x90BW#\x08\x00W#\x08\x00'
>>> b.hex()
'1402010020000000031890425723080057230800'

If you want the specific format, you could split it up and add the "\x" in front.

>>> h = b.hex()
>>> print("".join([f"\\x{h[i:i+2]}" for i,_ in enumerate(h[::2])]))
\x14\x40\x02\x20\x01\x10\x00\x02\x20\x00\x00\x00\x00\x00\x00\x00\x03\x31\x18\x89

thanks for your help.