Nrthugu

I really tried to look for this doubt in many ways but maybe since I never worked with binary files before I don't have any idea what the keywords to search similar things to help me out. That's why I am asking here.

So, I have a file:

path = 'myPath/file.pd0'

in_file = open(path, "rb")

read_file = in_file.read()

type(read_file) 

when I try to check what is inside read_file I get:

b'x7fx7fxccx05x00x0f$x00`x00xa2x00$x02xe6x02xa8x03xd0x032x04dx04x96x04xa6x04xe0x04'

The type of read_file is bytes. When I try to use struct since it is the function people suggest I get the following error:

import struct

struct.unpack('hhl', read_file[0:30])



error: unpack requires a buffer of 16 bytes

No matter what fmt I get unpack requires a buffer of n bytes.

The file structure that I am trying to read is defined as follow:

HEADER
(6 BYTES + [2 x No. OF DATA TYPES])

FIXED LEADER DATA
(60 BYTES)

VARIABLE LEADER DATA
(66 BYTES)

CORRELATION MAGNITUDE
(2 BYTES + 4 BYTES PER DEPTH CELL)

Any idea how I could start reading these bytes using struct or something similar in python?

Thank you

unpack() expects bytes which are the exact length of the format described by its first argument.
The format string 'hhl' describes data of 16 bytes (on your machine - see below), so you must pass a byte string of 16.
If you want to parse only part of the bytes, you can do this:

fmt = 'hhl'

size = struct.calcsize(fmt)

struct.unpack(fmt, data[:size])

Additionally, your format string doesn't have a byte order, size and alignment specifier. It is assumed to be "native" by default. This means your code is system-dependent, which is probably not what you want for parsing a file format.
You might need different alignments for different parts of the file.