2.2 features

[Skip Montanaro]

    >> I guess I'm confused, but if isinstance(x, type) is true isn't
    >> issubclass(x.__class__, type) also true?

    Guido> You are indeed confused. :)

    Guido> Seems you confuse isinstance(x, y) with issubclass(x, y).  These
    Guido> are very different.  x in y can map to at most one of these (for
    Guido> y a type object).

Here's an example that makes concrete what I was thinking:

    >>> class Foo:
    ...   pass
    ... 
    >>> class Bar(Foo):
    ...   pass
    ... 
    >>> x = Bar()
    >>> isinstance(x, Foo)
    1
    >>> issubclass(Bar, Foo)
    1
    >>> issubclass(x.__class__, Foo)
    1

Why can't the "in" operator grok all three of these possibilities?

    x in Foo                         same as isinstance(x, Foo)
    Bar in Foo                       same as issubclass(Bar, Foo)
    x.__class__ in Foo               same as issubclass(x.__class__, Foo)

I assume x can't be both an instance and a class at the same time.

Skip