Writing integers in binary to file in python

Question

How can I write integers to a file in binary in Python 3?

For example, I want to write 6277101735386680763835789423176059013767194773182842284081 to a file in binary in exactly 24 bytes (unsigned, I will only be working with positive integers). How can I do this? I tried the following:

struct.pack("i", 6277101735386680763835789423176059013767194773182842284081)

This results in

ValueError: cannot fit 'int' into an index-sized integer

I have tried the same with some other formats ("l", "Q"), but those also result in errors:

struct.error: argument out of range

If I could convert the integer to exactly 24 bytes, I would be able to write to the file, since I know how to do that. However, I can't seem to convert the integers into bytes.

Also, how do I make sure that exactly 24 bytes are written per integer? I will also be writing smaller numbers (1000, 2000, 1598754, 12), however those should also take 24 bytes.

And how can I read the integers again from the file afterwards?

Answer 1

With Python 3 you can do the following:

i = 6277101735386680763835789423176059013767194773182842284081
with open('out.bin', 'wb') as file:
    file.write((i).to_bytes(24, byteorder='big', signed=False))

with open('out.bin', 'rb') as file:
    j = int.from_bytes(file.read(), byteorder='big')

print(j)

Output:

$ python3 tiny.py
6277101735386680763835789423176059013767194773182842284081

Answer 2

You can extract the least significant byte with

  x = value & 0xFF

and you can remove that byte from the number with

  value = value >> 8

Repeating this procedure 24 times will give you the bytes.

You can also speed up this process by taking out more bytes, for example with

 x = value & 0xFFFFFFFF
 value = value >> 32

you're processing 4 bytes at a time.