Reading integers from binary file in Python

Manuel_28802 0
Viewed 168239

I'm trying to read a BMP file in Python. I know the first two bytes indicate the BMP firm. The next 4 bytes are the file size. When I execute:

fin = open("hi.bmp", "rb")
firm = fin.read(2)  
file_size = int(fin.read(4))

I get:

ValueError: invalid literal for int() with base 10: 'F#\x13'

What I want to do is reading those four bytes as an integer, but it seems Python is reading them as characters and returning a string, which cannot be converted to an integer. How can I do this correctly?

integer binary file python
7 Answers

The read method returns a sequence of bytes as a string. To convert from a string byte-sequence to binary data, use the built-in struct module: http://docs.python.org/library/struct.html. 

import struct

print(struct.unpack('i', fin.read(4)))

Note that unpack always returns a tuple, so struct.unpack('i', fin.read(4))[0] gives the integer value that you are after.

You should probably use the format string '<i' (< is a modifier that indicates little-endian byte-order and standard size and alignment - the default is to use the platform's byte ordering, size and alignment). According to the BMP format spec, the bytes should be written in Intel/little-endian byte order.

user_35710

Except struct you can also use array module

import array
values = array.array('l') # array of long integers
values.read(fin, 1) # read 1 integer
file_size  = values[0]
user_1081300

As you are reading the binary file, you need to unpack it into a integer, so use struct module for that

import struct
fin = open("hi.bmp", "rb")
firm = fin.read(2)  
file_size, = struct.unpack("i",fin.read(4))
user_69460

Here's a late solution but I though it might help.

fin = open("hi.bmp", "rb")
firm = fin.read(2)
file_size = 0
for _ in range(4):  
    (file_size << 8) += ord(fin.read(1))
user_107113110
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