从超类对象初始化子类实例

Question

The documentation for logging.Logger states:

Note that Loggers are never instantiated directly, but always through the module-level function logging.getLogger(name)

This leads me to the following question (for which I'll switch to some Foo class instead of logging.Logger , since I'd like to ask about the principle in general).

---

Suppose there's some class Foo , with several attributes and methods:

class Foo:
    def method_0(self..):
        ...

    def method_1(self..):
        ...

    def method_n(self..):
        ...

I'd like to subclass Foo , and specialize only a couple of these methods.

class SubFoo(Foo):
    def method_1(self, ...):
        ...
        # Or super - not the point of the question.
        Foo.method_1(self, ...)

    ...

The problem is that Foo doesn't have a constructor available, only the means to create an object of type Foo , say via getFoo . The " Foo " part of SubFoo needs to be an "approved" Foo object, and I don't know how to force SubFoo s Foo part to be made the "approved" object.

Note

There are obviously several workarounds to subclassing Foo , eg, composition, monkey-patching, clever games with __dict__ / gettatr , etc. Notwithstanding, this question is about subclassing.

Answer 1

You have misunderstood inheritance. To create an instance of SubFoo, you just have to instantiate it like you would any other class: SubFoo() .

Python doesn't have constructors, and in any case a method not defined on a subclass will automatically inherit the superclass version without you needing to do anything.