How can we override the constructor while still using the `class` keyword? [duplicate]

Question

Two notable ways to create a class are shown below:

class Klass:
    pass

Klass = type("Klass", tuple(), dict())

I would like to override the constructor (__call__) while still using the class keyword instead of doing something else, like directly calling type. I really do want to override (__call__), not __init__

My failed attempts are shown below:

Attempt 1

class Foo:
    @classmethod
    def __call__(*args):
        print("arr har")
        return super(type(args[0]), args[0]).__call__(*args)

instance = Foo()
# did not print "arr har"

Attempt 2

class BarMeta(type):
    def __call__(*args):
        print("hello world")
        return super(type(args[0]), args[0]).__call__(*args[1:])

Attempt 2A

class Bar:
    __metaclass__ = BarMeta

instance = Bar()
# did not print "hello world"

Attempt 2B

Baz = BarMeta("Baz", tuple(), dict())
instance = Baz()
# Did print "hello world," but we weren't able to use the `class` keyword to create `Baz`

But __call__ is not he constructor, __new__ is. If you want to override __call__ all you have to do is override __call__ — DeepSpace
– DeepSpace, Commented Oct 27, 2019 at 12:06
Attempt 2A would work if you were using python 2. In python 3, the syntax for defining a metaclass is class Bar(metaclass=BarMeta):. — Aran-Fey
– Aran-Fey, Commented Oct 27, 2019 at 12:08
@DeepSpace But __call__ calls __new__, not the other way around. instance = Klass() is the same as instance = Klass.__call__() or instance = type(Klass).__call__(Klass) — Toothpick Anemone
– Toothpick Anemone, Commented Oct 27, 2019 at 12:09
Also, I (and apparently at least 2 other people) had a hard time understanding your question. It would've been a good idea to start by explicitly stating what you want to do, i.e. "I want Bar() to print 'arr har'". — Aran-Fey
– Aran-Fey, Commented Oct 27, 2019 at 12:10
@DeepSpace type.__call__ is really the only example. repl.it/repls/EvenSandyPress — Aran-Fey
– Aran-Fey, Commented Oct 27, 2019 at 13:03

Toothpick Anemone · Accepted Answer · 2019-10-27 12:12:50Z

0

All credit does to Aran-Fey, who posted the answer as a comment, instead of as an answer:

class BarMeta(type):
    def __call__(*args):
        print("hello world")
        return super(type(args[0]), args[0]).__call__(*args[1:])    

class Bar(metaclass=BarMeta):
    pass

instance = Bar()

answered Oct 27, 2019 at 12:12

Toothpick Anemone

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