包装 python 超类的所有方法

Question

Is there a way to wrap all methods of a superclass, if I can't change its code?

As a minimal working example, consider this base class Base , which has many methods that return a new instance of itself, and the descendent class Child

class Base:
    
    def __init__(self, val):
        self.val = val
        
    def newinst_addseven(self):
        return Base(self.val + 7)
    
    def newinst_timestwo(self):
        return Base(self.val * 2)
    
    # ...

class Child(Base):
    
    @property
    def sqrt(self):
        return math.sqrt(self.val)

The issue here is that calling childinstance.newinst_addseven() returns an instance of Base , instead of Child .

Is there a way to wrap the Base class's methods to force a return value of the type Child ?

With something like this wrapper:

def force_child_i(result):
    """Turn Base instance into Child instance."""
    if type(result) is Base:
        return Child(result.val)
    return result

def force_child_f(fun):
    """Turn from Base- to Child-instance-returning function."""
    def wrapper(*args, **kwargs):
        result = fun(*args, **kwargs)
        return force_child_i(result)
    return wrapper

Many thanks!

---

PS: What I currently do, is look at Base 's source code and add the methods to Child directly, which is not very mainainable:

Child.newinst_addseven = force_child_f(Base.newinst_addseven)
Child.newinst_timestwo = force_child_f(Base.newinst_timestwo)

Answer 1

One option is to use a metaclass:

class ChildMeta(type):
    def __new__(cls, name, bases, dct):
        child = super().__new__(cls, name, bases, dct)
        for base in bases:
            for field_name, field in base.__dict__.items():
                if callable(field):
                    setattr(child, field_name, force_child(field))
        return child


class Child(Base, metaclass=ChildMeta):
    pass

It will automatically wrap all the Base s methods with your force_child decorator.