I had a problem: using Python 2.7, it was not possible to create a subprocess using
subprocess.Popen([.......], close_fds=True, stdout=subprocess.PIPE, ...)
         
        This issue is fixed on Python 3.7+ by default
This is definatly a tricky hack: answer is to iterate through already opened file descriptors, just before using the subprocess module.
def _hack_windows_subprocess():
    """HACK: python 2.7 file descriptors.
    This magic hack fixes https://bugs.python.org/issue19575
    by adding HANDLE_FLAG_INHERIT to all already opened file descriptors.
    """
    # See https://github.com/secdev/scapy/issues/1136
    import stat
    from ctypes import windll, wintypes
    from msvcrt import get_osfhandle
    HANDLE_FLAG_INHERIT = 0x00000001
    for fd in range(100):
        try:
            s = os.fstat(fd)
        except:
            continue
        if stat.S_ISREG(s.st_mode):
            handle = wintypes.HANDLE(get_osfhandle(fd))
            mask   = wintypes.DWORD(HANDLE_FLAG_INHERIT)
            flags  = wintypes.DWORD(0)
            windll.kernel32.SetHandleInformation(handle, mask, flags)
Here is a sample that would crash without it:
import os, subprocess
f = open("a.txt", "w")
subprocess.Popen(["cmd"], stdin=subprocess.PIPE, stdout=subprocess.PIPE, stderr=subprocess.STDOUT)
f.close()
os.remove(f.name)
Traceback (most recent call last):
File "stdin", line 1, in module
WindowsError: [Error 32] Le processus ne peut pas accÚder au fichier car ce fichier est utilisÚ par un autre processus: 'a.txt'
Now with the fix:
import os, subprocess
f = open("a.txt", "w")
_hack_windows_subprocess()
subprocess.Popen(["cmd"], stdin=subprocess.PIPE, stdout=subprocess.PIPE, stderr=subprocess.STDOUT)
f.close()
os.remove(f.name)
Works.
Hope i helped