Armin Rigo added the comment:
(B9) CPython 3.5.2: this ``nonlocal`` seems not to have a reasonable
effect (note that if we use a different name instead of ``__class__``,
this example correctly complain that there is no binding in the outer
scope of ``Y``)::
class Y:
class X:
nonlocal __class__
__class__ = 42
print(locals()['__class__']) # 42
print(__class__) # but this is a NameError
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<http://bugs.python.org/issue28884>
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