Class __init__ behaviour

[Thomas Bartkus]

If I insert an __init__ method in my own class definition, it is incumbent
upon me to call the __init__ of any declared ancester to my new class object
because my __init__ will override that of any ancester I declare in the
header. If I fail to call the ancesters __init__, then it won't happen.
The ancester object won't be initialized.

But

If I *don't* insert my own __init__ in my new class, then any declared
ancester __init__ will automatically run because I haven't overridden the
ancesters __init__ method with my own.

Did I get that straight?
Thomas Bartkus

 

[Diez B. Roggisch]

If I insert an __init__ method in my own class definition, it is incumbent
upon me to call the __init__ of any declared ancester to my new class object
because my __init__ will override that of any ancester I declare in the
header. If I fail to call the ancesters __init__, then it won't happen.
The ancester object won't be initialized. Yes.


But

If I *don't* insert my own __init__ in my new class, then any declared
ancester __init__ will automatically run because I haven't overridden the
ancesters __init__ method with my own.

No. Only the __init__ of the leftmost ancestor is called. Of course if
that uses the super(..)-method the other constructors get called.

Diez

 

[Lou Pecora]

If I insert an __init__ method in my own class definition, it is incumbent
upon me to call the __init__ of any declared ancester to my new class object
because my __init__ will override that of any ancester I declare in the
header. If I fail to call the ancesters __init__, then it won't happen.
The ancester object won't be initialized.

But

If I *don't* insert my own __init__ in my new class, then any declared
ancester __init__ will automatically run because I haven't overridden the
ancesters __init__ method with my own.

Did I get that straight?
Thomas Bartkus

Sounds right to me. That's how I use it.

-- Lou Pecora (my views are my own) REMOVE THIS to email me.