How can I get integer values from binary file

Question

I have some files they look like this with hex/octal dump:

hexdump file:

0000000 0002 0000 0c3e 0000 0ac5 0000          
000000c

od file:

0000000 000002 000000 006076 000000 005305 000000
0000014

I dont know really what this "000002" and "0000014" is good for but, I know I need this "006076" and "005305" as interger value, so I did this in python2:

hex(006076)
'0xc3e'

int(0xc3e)
3134

I don't know really how I should open/and read this file/values in python2...

simple open/read the file looks like this:

x = open("file", "rb")
x.read()
'\x02\x00\x00\x00>\x0c\x00\x00\xc5\n\x00\x00'

0000000 000002 is one value, by the way, but try to open the file, at least, and read it — OneCricketeer
– OneCricketeer, Commented May 26, 2017 at 13:26
Just because the file contents are binary strings, doesn't mean you need to open the file as binary — OneCricketeer
– OneCricketeer, Commented May 26, 2017 at 13:35
You're getting completely different contents when you're opening the file in Python. Are you really sure you're opening the same file that you hexdumped earlier? Either you're opening a different file, or something has changed the file in the meantime. — Dolda2000
– Dolda2000, Commented May 26, 2017 at 13:41

chepner · Accepted Answer · 2017-05-26 13:56:06Z

3

You want to use struct.unpack.

>>> import struct
>>> struct.unpack("4x H 2x H 2x", '\x02\x00\x00\x00>\x0c\x00\x00\xc5\n\x00\x00')
(3134, 2757)

The format string used as the first argument tells unpack to ignore the first 4 bytes, then unpack the next two bytes as an integer, ignore 2 more bytes, unpack another 2-byte integer, and ignore the last 2 bytes. This assumes the byte string uses the same endianness as your machine's default; see the documentation for the struct module for more details.

answered May 26, 2017 at 13:56

chepner

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izoughe · Accepted Answer · 2017-05-26 14:23:28Z

0

The functions hex(int) and oct(int) take an integer in base 10 (decimal) and convert it into a string that looks like a number in base 8 or base 16 (hex or octal). The function int() just takes a string that looks like a number and converts it from a string into an integer in base 10.

You should be able to use the int() function to do this with the format int([string representing number], [base of number]). Here's how it works:

In a base n number system, the value of the number is (1st digit) * n^0 + (2nd digit) * n^1 + (3rd digit) ^ n^2 + .... For example, if you have the number 472 in octal (base 8), you could calculate the value like so:

2 * 8^0  +  7 * 8^1  +  4 * 8^2
2 * (1)  +  7 * (8)  +  4 * (64)
2 + 56 + 256
314

So 472 in octal is equivalent to 314 in decimal. To make a function out of this, I'd do something like this:

def convertToDecimal(number, baseOfNumber):
    counter = 0
    for c in reversed(number)
        newNumber += (c * (baseOfNumber ^ counter))
        counter += 1
    return newNumber

Hope this helps!

edited May 26, 2017 at 14:23

answered May 26, 2017 at 13:50

izoughe

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3 Comments

PM 2Ring Over a year ago

FWIW, the int constructor takes an optional base arg. Try this: int('472',8).

izoughe Over a year ago

Well, that's a lot simpler than my answer. Thank you!

izoughe Over a year ago

Edited the answer to reflect that.

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