Making a class with empty name using type

According to Python's documentation for the built-in function type:

> class type(name, bases, dict, **kwds)
>
> [...]
>
> With three arguments, return a new type object. This is essentially a dynamic form of the class statement. The name string is the class name and becomes the __name__ attribute.

I have realized that the name string might as well be empty, and the function invocation would still work:

>>> type('', (), {})
<class '__main__.'>

Is there any issue that might arise from doing it?

Here is one thing which won't work correctly: pickle.

This works, because it has a valid identifier for class name:

>>> Good = type('Good', (), {})
>>> setattr(__main__, 'Good', Good)
>>> good = Good()
>>> pickle.loads(pickle.dumps(good))
<__main__.Good object at 0x7ff0d8173160>

This does not work, because the class name is empty:

>>> Bad = type('', (), {})
>>> setattr(__main__, '', Bad)
>>> bad = Bad()
>>> pickle.loads(pickle.dumps(bad))
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
_pickle.UnpicklingError: pickle data was truncated

I am sure there are other places where one could encounter problems. Using the "normal" way to define types does not allow class names like '' or '77'. It is correct for any function to assume that.

When using type to create a new type, one should make sure that the same restrictions are followed, instead of assuming that libraries are going to handle invalid identifiers as if they were valid.