Python.. convert bin/hex file to txt file

Question

I want to convert a bin file to txt file in python.

with open("atb.bin", "rb") as file:
 data = file.read(8)
datastring = str(data)
print(datastring)
print(' '.join(str(ord(c)) for c in datastring))

The output I get is

b'\x14\x12\x1c\x1a#\x00-d'
98 39 92 120 49 52 92 120 49 50 92 120 49 99 92 120 49 97 35 92 120 48 48 45 100 39

Now, how do I convert this to decimal and store in a txt file?

• Jean-François Fabre: don't convert your bytes object to string... python 2 or python 3?
• Aj Mv: Its Python 3.Can you please suggest a solution?
• majin: Possible duplicate of Reading a binary file with python

Answer 1

Your main mistake is doing:

datastring = str(data)

As it converts the representation of the data object to string, with b prefix and quotes, and escaping... wrong.

read your file as you're doing:

with open("atb.bin", "rb") as file:
 data = file.read(8)

now data is a bytes object, no need for ord in python 3, values are already integer. Now open a text file and dump the values (converting integers to string, decimal):

with open("out.txt", "w") as f:
 f.write(" ".join(map(str,data))):
 f.write("\n")

In python 2 you'd need to get the character code then convert to string:

 f.write(" ".join([str(ord(c)) for c in data]))

Answer 2

Your Python converts into text, the text representation of the 8 characters in the file.

Hence, instead of

print(' '.join(str(ord(c)) for c in datastring))

you should have put

print(' '.join(str(ord(c)) for c in data))

which you can then write to a file using standard techniques.

ie. (Python 2)

>>> data=b'\x14\x12\x1c\x1a#\x00-d'
>>> print(' '.join(str(ord(c)) for c in data))
20 18 28 26 35 0 45 100

(Python 3)

>>> data=b'\x14\x12\x1c\x1a#\x00-d'
>>> print(' '.join(str(c) for c in data))
20 18 28 26 35 0 45 100

• Aj Mv: Can we use struct.unpack() in this case? In which scenarios the unpack is useful?