[FIXED] How to alias a Python constructor?

Issue

I read here and here that Python methods can be aliased using =.

However, when I try it on the __init__ method, it doesn't seem to work.

class MyClass:
    def __init__(self):
        print("Hi mum!")
    
    new_name = __init__

a = MyClass()
b = MyClass.new_name()

The b = MyClass.new_name() causes this error:

TypeError: __init__() missing 1 required positional argument: 'self'

Why does this not work? And how should I alias __init__ correctly?

Solution

__init__ is not a constructor. __new__ is. __init__ is an instance method that performs initialization on an instance after it is constructed by the static method __new__. The reason why both __new__ and __init__ are called when you call the class with MyClass() is because it is what the __call__ method of the type, or metaclass, of MyClass, does.

So if you want to properly alias the constructor of a class you can alias the __call__ method of its metaclass:

class AliasedConstructor(type):
    new_name = type.__call__

class MyClass(metaclass=AliasedConstructor):
    def __init__(self):
        print("Hi mum!")

MyClass.new_name()

Alternatively, you can make the alias a classmethod descriptor that binds type.__call__ to the current class:

class MyClass:
    def __init__(self):
        print("Hi mum!")

    new_name = classmethod(type.__call__)

Both approaches would output:

Hi mum!

Answered By - blhsing