Richard Oudkerk <[email protected]> added the comment:
I checked the source in
c:/Program Files (x86)/Microsoft Visual Studio 10.0/VC/crt/src/open.c
and it seems that on Windows open() is more or less implemented as a wrapper of
sopen(..., ..., SH_DENYNO, ...).
So the only reason that trying to reopen a NamedTemporaryFile fails on Windows
is because when we reopen we need to use O_TEMPORARY.
The following works for unmodified python:
import os, tempfile
DATA = b"hello bob"
def temp_opener(name, flag, mode=0o777):
return os.open(name, flag | os.O_TEMPORARY, mode)
with tempfile.NamedTemporaryFile() as f:
f.write(DATA)
f.flush()
with open(f.name, "rb", opener=temp_opener) as f:
assert f.read() == DATA
assert not os.path.exists(f.name)
So maybe we should just define tempfile.opener().
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<http://bugs.python.org/issue14243>
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