如果定义了超类__new__，则不调用子类__init__

Question

See the following ipython session:

$ ipython --no-banner

In [1]: class foo(object):
   ...:     def __new__(self, *args, **kwargs):
   ...:         print 'in superclass __new__'
   ...:         return self
   ...: 
   ...: class bar(foo):
   ...:     def __init__(self, *args, **kwargs):
   ...:         print 'in subclass __init__'
   ...:         self.val = 42
   ...:         

In [2]: b = bar()
in superclass __new__

In [3]: b.val
---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
<ipython-input-4-c55d38c9d4d9> in <module>()
----> 1 b.val

AttributeError: type object 'bar' has no attribute 'val'

What is going on here? Why does defining __new__ in foo prevent the __init__ from running in bar ?

Answer 1

You are returning an invalid value from __new__ . Quoting the documentation :

If __new__() returns an instance of cls , then the new instance's __init__() method will be invoked ... If __new__() does not return an instance of cls , then the new instance's __init__() method will not be invoked.

Note that the first argument to __new__ is a class, not an object. __new__ 's job is to allocate the object, often by calling super().__new__ .

Try this:

class foo(object):
    def __new__(cls, *args, **kwargs):
        print 'in superclass __new__'
        return super(foo, cls).__new__(cls, *args, **kwargs)

class bar(foo):
    def __init__(self, *args, **kwargs):
        print 'in subclass __init__'
        self.val = 42

b = bar()
print b

Answer 2

From the datamodel documentation:

If __new__() does not return an instance of cls , then the new instance's __init__() method will not be invoked.

So, try again like this:

class foo(object):
    def __new__(cls, *args, **kwargs):
        print 'in superclass __new__'
        return super(foo, cls).__new__(cls, *args, **kwargs)

class bar(foo):
    def __init__(self, *args, **kwargs):
        print 'in subclass __init__'
        self.val = 42

Answer 3

Your __new__ function signature implies you don't quite understand what is going on. The first argument to __new__ will be the class , calling it self which is conventionally used for the instance should be a warning: __new__ is a constructor, the instance doesn't exist yet . So what happens is you are returning an object that isn't the same type as the class, (it is the class in fact, so it is of type type ) thus, __init__ is skipped! You want something like this:

In [4]: class Foo(object):
   ...:     def __new__(cls, *args, **kwargs):
   ...:         print 'in superclass __new__'
   ...:         return super(Foo, cls).__new__(cls, *args, **kwargs)
   ...:
   ...: class Bar(Foo):
   ...:     def __init__(self, *args, **kwargs):
   ...:         print 'in subclass __init__'
   ...:         self.val = 42
   ...:

In [5]: b = Bar()
in superclass __new__
in subclass __init__