Understanding / Using __init__ method of dict-CodePudding

I came across this idea how to make a Python class serializeable.

class FileItem(dict):
    def __init__(self, name):
        dict.__init__(self, name=name)

x = FileItem("test")
print(x)

{'name': 'test'}

This works fine, but I do not understand how. I thought dict.__init__ calls the __init__ method from the dict class to create a new instance and so I expected this to work:

x = dict.__init__(name="test")

This results in the following error:

TypeError: descriptor '__init__' of 'dict' object needs an argument

Why is this different to the example above?

CodePudding user response：

So, notice that you've not provided a second argument to __init__,

>>> x = dict.__init__(name="test")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: descriptor '__init__' of 'dict' object needs an argument

It require the instance, i.e self, as the first argument. Of course, if you create an appropriate instance and pass that, __init__ returns None:

>>> x = dict.__init__({}, name="test")
>>> print(x)
None

Because __init__ is not a constructor, it is an initializer. In Python these are two distinct steps in the object initialization process.

The "constructor" is __new__:

>>> dict.__new__(dict)
{}

Read about it in the Python data model docs.

Also, for the sake of decency, you probably shouldn't be sprinkling:

x = dict.__new__(dict)

In real code. But this is how the mechanics of it work. All of this happens in type.__call__, by the way. type is the parent class that creates classes, a metaclass. (It is also itself a type... type(type) is type just like type(dict) is type... )

>>> type.__call__(dict)
{}

CodePudding user response：

__new__ creates new instances but __init__ is used for initializing a newly created instance, so __init__ requires an instance to work with.

When you do dict.__init__(name='test') then this isn't operating on any instance, hence the error. With dict.__init__(self, name=name) on the other hand, you do pass the instance self as an argument and hence it works.

Typically you'd use super().__init__(name=name) which takes care of providing self as an argument and also works when you decide to changes base classes later on.

CodePudding user response：

The difference is the self property which isn't passed, the __init__ property applies its computations on the instance (self) passed in, if the __init__ function doesn't have any context, it throws an error
Have a great day, Lukas.