Why does os.fdopen() ignore "mode" argument?

Question

This piece of code when run on Python 2.7.16 and 3.8.3 produces different results:

import tempfile
import os

fd, lockfile = tempfile.mkstemp()
flags = os.O_RDWR | os.O_CREAT
mode = 'w+b'

fd = os.open(lockfile, flags)
fileobj = os.fdopen(fd, mode)

print(fileobj.mode)

os.remove(lockfile)

In 2.7 it prints w+b as expected but in 3.8 it prints rb+. Why wouldn't it be respecting the mode parameter in this way?

I've tried manually creating a file to discount tempfile differences but still get the same result.

I can't see anything obvious in the docs:

• https://docs.python.org/3.8/library/os.html#os.fdopen
• https://docs.python.org/2.7/library/os.html#os.fdopen

Running on MacOS 10.14.6

Answer 1

From the documentation of the built-in open function:

mode is an optional string that specifies the mode in which the file is opened.

When open is called with a file descriptor rather than a file path (or when you use the alias fdopen that requires a file descriptor), no file is opened. A Python file-like object that wraps the file descriptor is created and returned. You can't change the mode of an open file, so the mode argument is simply ignored.

Answer 2

I'm not sure if Python tracks this, but given your flags used in os.open, I'd say that rb+ is actually correct.

You called os.open with flags "read/write" and "create if not exist", but without "truncate" (O_TRUNC). This is exactly the difference between modes rb+ and wb+. Assume Python tracked your flags, it's the correct mode.