yves at zioup
May 14, 2008, 5:23 PM
Views: 708
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I want to create a subclass of 'file' but need to open the file with os.open
(because I want to open it in exclusive mode), and need an additional method.
Because I need an additional method, I truly need a object of my sublass.
If I do something like
class myFile(file):
def __new__(cls, filename):
import os
fd = os.open(filename, os.O_WRONLY | os.O_CREAT | os.O_EXCL)
return os.fdoen(fd, 'w')
def myMethod(self):
do_something
then x = myFile('somefilename') is of type file, not myFile, and therefore
does not have myMethod as a valid method.
Now if I try:
class myFile(file):
def __new__(cls, filename):
import os
fd = os.open(filename, os.O_WRONLY | os.O_CREAT | os.O_EXCL)
obj = file.__new__(cls)
return obj
def myMethod(self):
do_something
Then it fails, because the 'file' constructor needs a filename. I can provide
the filename, but it will then try to re-open that file, and even if I did
manage to create an object file, how do I connect the file descriptor created
by os.open to my object ?
Thanks.
Yves.
http://www.SollerS.ca
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How to subclass file
