type(socket) != SocketType

Irmen de Jong irmen at NOSPAMREMOVETHISxs4all.nl
Sat Jun 15 23:10:37 CEST 2002


Hello

I have to detect if a certain socket object is a regular socket,
or a M2Crypto SSL socket. I do this by checking type(sock),
and then checking if this is a regular socket type. Why this way?
because the systems that run my code do not necessarily have
M2Crypto installed, and thus I cannot compare the socket to
any class/type from M2Crypto, nor just go ahead and catch SSL
exceptions (simply because they are not defined!).

There is a problem, however.
With Python 2.2.1 on Windows, type(socket) != SocketType:

 >>> from socket import *
 >>> s=socket(AF_INET,SOCK_STREAM)
 >>> type(s)
<type 'instance'>
 >>> SocketType
<type '_socket.socket'>
 >>> type(s) is SocketTye
Traceback (most recent call last):
   File "<stdin>", line 1, in ?
NameError: name 'SocketTye' is not defined
 >>> type(s) is SocketType
0
 >>> type(s)== SocketType
0
 >>>


I find this most unfortunate, because the docs on the socket module
explicitly state that type(socket)=SocketType.

BTW on Linux, all is fine.

Any clues how to fix things on windows?

Thanks a lot,
Irmen




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