Python，如何将32位整数放入字节数组中

Question

I usually perform things like this in C++, but I'm using python to write a quick script and I've run into a wall.

If I have a binary list (or whatever python stores the result of an "fread" in). I can access the individual bytes in it with: buffer[0], buffer[1], etc.

I need to change the bytes [8-11] to hold a new 32-bit file-size (read: there's already a filesize there, I need to update it). In C++ I would just get a pointer to the location and cast it to store the integer, but with python I suddenly realized I have no idea how to do something like this.

How can I update 4 bytes in my buffer at a specific location to hold the value of an integer in python?

EDIT

I'm going to add more because I can't seem to figure it out from the solutions (though I can see they're on the right track).

First of all, I'm on python 2.4 (and can't upgrade, big corporation servers) - so that apparently limits my options. Sorry for not mentioning that earlier, I wasn't aware it had so many less features.

Secondly, let's make this ultra-simple.

Lets say I have a binary file named 'myfile.binary' with the five-byte contents '4C53535353' in hex - this equates to the ascii representations for letters "L and 4xS" being alone in the file.

If I do:

f = open('myfile.binary', 'rb')
contents = f.read(5)

contents should (from Sven Marnach's answer) hold a five-byte immutable string.

Using Python 2.4 facilities only, how could I change the 4 S's held in 'contents' to an arbitrary integer value? Ie give me a line of code that can make byte indices contents [1-4] contain the 32-bit integer 'myint' with value 12345678910.

Answer 1

What you need is this function:

struct.pack_into(fmt, buffer, offset, v1, v2, ...)

It's documented at http://docs.python.org/library/struct.html near the top.

Example code:

import struct
import ctypes

data=ctypes.create_string_buffer(10)
struct.pack_into(">i", data, 5, 0x12345678)
print list(data)

Similar posting: Python: How to pack different types of data into a string buffer using struct.pack_into

EDIT: Added a Python 2.4 compatible example:

import struct

f=open('myfile.binary', 'rb')
contents=f.read(5)
data=list(contents)
data[0:4]=struct.pack(">i", 0x12345678)
print data

Answer 2

Have a look at Struct module. You need pack function.

EDIT:

The code:

import struct

s = "LSSSS" # your string
s = s[0] + struct.pack('<I', 1234567891) # note "shorter" constant than in your example
print s

Output:

L╙☻ЦI

struct.pack should be available in Python2.4.

Your number "12345678910" cannot be packed into 4 bytes, I shortened it a bit.

Answer 3

The result of file.read() is a string in Python, and it is immutable. Depending on the context of the task you are trying to accomplish, there are different solutions to the problem.

One is using the array module : Read the file directly as an array of 32-bit integers. You can modify this array and write it back to the file.

with open("filename") as f:
    f.seek(0, 2)
    size = f.tell()
    f.seek(0)
    data = array.array("i")
    assert data.itemsize == 4
    data.fromfile(f, size // 4)
data[2] = new_value
# use data.tofile(g) to write the data back to a new file g

Answer 4

You could install the numpy module, which is often used for scientific computing.

read_data = numpy.fromfile(file=id, dtype=numpy.uint32)

Then access the data at the desired location and make your changes.

Answer 5

The following is just a demonstration for you to understand what really happens when the four bytes are converted into an integer. Suppose you have a number: 15213

Decimal: 15213
Binary: 0011 1011 0110 1101
Hex: 3 B 6 D

On little-endian systems (ie x86 machines), this number can be represented using a length-4 bytearray as: b"\\x6d\\x3b\\x00\\x00" or b"m;\\x00\\x00" when you print it on the screen, to convert the four bytes into an integer, we simply do a bit of base conversion, which in this case, is:

sum(n*(256**i) for i,n in enumerate(b"\x6d\x3b\x00\x00"))

This gives you the result: 15213