Ask Solem <a...@opera.com> added the comment:
Maybe surprising but not so weird if you think about what happens
behind the scenes.
When you do
>>> x = man.list()
>>> x.append({})
You send an empty dict to the manager to be appended to x
when do:
>>> x[0]
{}
you receive a local copy of the empty dict from the manager process.
So this:
>>> x[0]["a"] = 5
will only modify the local copy.
What you would have to do is:
>>> x.append({})
>>> t = x[0]
>>> t["a"] = 5
>>> x[0] = t
This will not be atomic of course, so this may be something
to take into account.
What maybe could be supported is something like:
>>> x[0] = manager.dict()
>>>x[0]["foo"] = "bar"
but otherwise I wouldn't consider this a bug.
----------
nosy: +asksol
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Python tracker <rep...@bugs.python.org>
<http://bugs.python.org/issue9801>
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