Implementation of type(name, bases, dict_)

Asked Feb 05 '18 at 06:50

Active Feb 05 '18 at 06:50

Viewed 127 times

I'm trying to understand how type works, specifically the three-argument form. How would this look?

Here's my shot, but this is a TypeError.

def type_(name, bases, dict_):
    base = next(base for base in bases if hasattr(base, '__new__'))
    return super(type(base)).__new__(type(base))

type('A', (int, object,), {}) # __main__.A
type_('A', (int, object,), {})

TypeError: super.__new__(type): type is not a subtype of super

asked Feb 05 '18 at 06:50

Hatshepsut

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What are you trying to do with `super` here? – juanpa.arrivillaga Feb 05 '18 at 06:56
`type` is a class, like `list`, or `str`. It creates class-objects. It is a [metaclass](https://docs.python.org/3/reference/datamodel.html#metaclasses) – juanpa.arrivillaga Feb 05 '18 at 07:30
@juanpa.arrivillaga Yes type is a class. Is it possible to show what `type()` does, using a function or a class? – Hatshepsut Feb 05 '18 at 07:52
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[There is a very nice explanation on metaclasses and type](https://stackoverflow.com/a/6581949/5741172), Have a look. – Sohaib Farooqi Feb 05 '18 at 08:01

Implementation of type(name, bases, dict_)

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