On 2008-05-07 13:37, Amaury Forgeot d'Arc wrote:
Hello,
2008/5/7 Sjoerd Mullender <[EMAIL PROTECTED]>:
Why does sock.close() not actually close sock?
If I run the code
import socket
sock = socket.socket()
...
sock.close()
I would expect that a system call is done to actually close the socket and
free the file descriptor. But that does not happen. Look at the code in
socket.py. It merely replaces the socket instance with a dummy instance so
that all subsequent calls on the sock object fail, but it does nothing else!
It does close the socket:
In socket.py, when self._sock is replaced, its __del__ method will be called.
This __del__ is implemented in C, in socketmodule.c:
static void
sock_dealloc(PySocketSockObject *s)
{
if (s->sock_fd != -1)
(void) SOCKETCLOSE(s->sock_fd);
Py_TYPE(s)->tp_free((PyObject *)s);
}
Of course, if you call sock.dup() or sock.makefile(),
there is another reference to the underlying _sock, and you must
close() all these objects.
I have to question the design of this. When I close() an object I
expect it to be closed there and then and not at some indeterminate
later time (well, it is determinate when you're fully aware of all
references, but often you aren't--trust me, I understand reference
counting).
Then there also seems to be a bug in imaplib.IMAP4_SSL since in its
shutdown method it closes the socket but leaves the sslobj untouched. I
assume that that object keeps a reference to the socket.
But as I said, I expect the close() to actually close.
--
Sjoerd Mullender
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