Kent Johnson wrote: > Peter Hansen wrote: >> That will not work, and you will get data loss, as Jeremy points out. >> > Can you explain why not? self.data is still bound to the same list as x. > At least if the execution sequence is x = self.data > self.data.append(a_piece_of_data) > self.data = []
Ah, since the entire list is being returned, you appear to be correct. Interesting... this means the OP's code is actually appending things to a list, over and over (presumably), then returning a reference to that list and rebinding the internal variable to a new list. If another thread calls on_received() and causes new data to be appended to "the list" between those two statements, then it will show up in the returned list (rather magically, at least to my way of looking at it) and will not in fact be lost. Good catch Kent. :-) -Peter -- http://mail.python.org/mailman/listinfo/python-list
