Determine the defining class from within a method

Question

The following Python 3.5 code:

class Base(object):
    def __init__(self):
        print("My type is", type(self))

class Derived(Base):
    def __init__(self):
        super().__init__()
        print("My type is", type(self))

d = Derived()

prints:

My type is <class '__main__.Derived'>
My type is <class '__main__.Derived'>

I would like to know, inside each __init__(), the class where the method was defined, not the deriving class. So I would get the following print:

My type is <class '__main__.Base'>
My type is <class '__main__.Derived'>

Related: http://stackoverflow.com/questions/961048/get-class-that-defined-method. — alecxe, Jan 17 '16 at 13:59
@alecxe I saw that link, it's not relevant to my question since 1) both Base and Derived have __init__ method so relaying on the name is not helpful. 2) I would like to know the defining class from within the __init__ method, and for any level of inheritance. — Periodic Maintenance, Jan 17 '16 at 14:10

Mike Müller · Accepted Answer · 2016-01-17T14:26:06.983

Solution 1

Use super().__thisclass__:

class Base(object):
    def __init__(self):
        print("My type is", super().__thisclass__)

class Derived(Base):
    def __init__(self):
        super().__init__()
        print("My type is", super().__thisclass__)

d = Derived()

My type is <class '__main__.Base'>
My type is <class '__main__.Derived'>

Solution 2

Not as elegant but hard-wiring the class works:

class Base(object):
    def __init__(self):
        print("My type is", Base)

class Derived(Base):
    def __init__(self):
        super().__init__()
        print("My type is", Derived)

d = Derived()

Output:

My type is <class '__main__.Base'>
My type is <class '__main__.Derived'>

Indeed super().__thisclass__ (solution 1) does the trick.This is exactly what I needed. — Periodic Maintenance, Jan 17 '16 at 14:32

Determine the defining class from within a method

1 Answers1

Solution 1

Solution 2