Team, Are we supposed to be able to have our own class dictionary in python 3?
If we currently cannot -- do we want to be able to? That we can have out own class dictionary in python 3 is strongly implied in the following at https://www.python.org/dev/peps/pep-3115/ where it says: """ # The metaclass invocation def __new__(cls, name, bases, classdict): # Note that we replace the classdict with a regular # dict before passing it to the superclass, so that we # don't continue to record member names after the class # has been created. result = type.__new__(cls, name, bases, dict(classdict)) result.member_names = classdict.member_names return result """ I don't understand this. As far as I can tell, no matter what class dictionary you pass into `type.__new__` it creates a copy of it. Am I missing something? Is this supposed to work? Is the documentation wrong? Thanks, Joy Diamond. Program that shows that the class dictionary created is not what we pass in --- Shows the actual symbol table is `dict` not `SymbolTable` class SymbolTable(dict): pass members = SymbolTable(a = 1) X = type('X', ((object,)), members) members['b'] = 2 print('X.a: {}'.format(X.a)) try: print('X.b: {}'.format(X.b)) except AttributeError as e: print('X.b: does not exist') # # Get the actual symbol table of `X`, bypassing the mapping proxy. # X__symbol_table = __import__('gc').get_referents(X.__dict__)[0] print('The type of the actual symbol table of X is: {} with keys: {}'.format( type(X__symbol_table), X__symbol_table.keys())) # Prints out # X.a: 1 # X.b: does not exist # The type of the actual symbol table of X is: <class 'dict'> with keys: dict_keys(['a', '__module__', '__dict__', '__weakref__', '__doc__']) <https://www.avast.com/sig-email?utm_medium=email&utm_source=link&utm_campaign=sig-email&utm_content=webmail&utm_term=icon> Virus-free. www.avast.com <https://www.avast.com/sig-email?utm_medium=email&utm_source=link&utm_campaign=sig-email&utm_content=webmail&utm_term=link> <#DAB4FAD8-2DD7-40BB-A1B8-4E2AA1F9FDF2>
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