Can I set the umask for tempfile.NamedTemporaryFile in python?

Question

In Python (tried this in 2.7 and below) it looks like a file created using tempfile.NamedTemporaryFile doesn't seem to obey the umask directive:

import os, tempfile
os.umask(022)
f1 = open ("goodfile", "w")
f2 = tempfile.NamedTemporaryFile(dir='.')
f2.name

Out[33]: '/Users/foo/tmp4zK9Fe'

ls -l
-rw-------  1 foo  foo  0 May 10 13:29 /Users/foo/tmp4zK9Fe
-rw-r--r--  1 foo  foo  0 May 10 13:28 /Users/foo/goodfile

Any idea why NamedTemporaryFile won't pick up the umask? Is there any way to do this during file creation?

I can always workaround this with os.chmod(), but I was hoping for something that did the right thing during file creation.

Answer 1

This is a security feature. The NamedTemporaryFile is always created with mode 0600 , hardcoded at tempfile.py , line 235 , because it is private to your process until you open it up with chmod . There is no constructor argument to change this behavior.

Answer 2

In case it might help someone, I wanted to do more or less the same thing, here is the code I have used:

import os
from tempfile import NamedTemporaryFile

def UmaskNamedTemporaryFile(*args, **kargs):
    fdesc = NamedTemporaryFile(*args, **kargs)
    # we need to set umask to get its current value. As noted
    # by Florian Brucker (comment), this is a potential security
    # issue, as it affects all the threads. Considering that it is
    # less a problem to create a file with permissions 000 than 666,
    # we use 666 as the umask temporary value.
    umask = os.umask(0o666)
    os.umask(umask)
    os.chmod(fdesc.name, 0o666 & ~umask)
    return fdesc