从二进制文件解包 4 字节无符号整数给出 struct.error：解包需要 4 字节的缓冲区

Question

This is a repeated question but I couldn't find answer

I am reading a binary file in following format (just showing one of the 3000 lines in hex format as shown in sublime text):

0009 7f71 0009 b87b 0009 f24b 000a 2ce2

I want to read it as tuple of 4-byte unsigned integers

if filename:
    with open(filename, mode='rb') as file:
        fileData = file.read()
        unsignedData = struct.unpack('I', fileData )

However I get following error for last line in above code:

struct.error: unpack requires a buffer of 4 bytes

How to fix this?

Answer 1

Decoding a few integers

The io.RawIOBase.read will usually return a lot more than 4 bytes (it's limited by the system call and/or file size).

On the other hand, the buffer's size in bytes must match the size required by the format string to struct.unpack .

The overall data structure of your file is not clear, but to, for example, read 4 unsigned 32-bit integers encoded in little-endian (data you provided), you should slice the buffer:

unsignedData = struct.unpack('4I', fileData[:16])

Decoding a stream

In case you need to decode an arbitrarily long stream of integers from your file, there are a few options, depending on the expected data length.

Short stream

with open(filename, mode='rb') as fp:
    fileData = fp.read()
    n, r = divmod(len(fileData), struct.calcsize('I'))
    assert r == 0, "Data length not a multiple of int size"
    unsignedData = struct.unpack('I' * n, fileData)

Long stream

You could use struct.iter_unpack , but it probably makes more sense to keep the large int data in an array.array or numpy.array anyway.

Here's an example for a NumPy array:

import numpy
data = numpy.fromfile(filename, dtype='uint32')

I'll leave the loading into Python's homogeneous data array as an exercise to the reader (hint: array.fromfile ).