Call Python object's variable as a function without passing the calling object

Question

I have a structure something like this:

def foobar():
    print('FOOBAR!')

class SampleClass:
    foo = foobar
    def printfunc(self):
        self.foo()

This doesn't work because the original foobar function can't handle self being passed to it -- it wasn't part of any class or object to begin with. Python won't let me add the @staticmethod decorator either.

I have no control over the definition of foobar, and I may have to override the value of foo in subclasses.

How can I call foo, without passing the object calling it?

Answer 1

Decorators are plain functions, you should be able to call staticmethod(foobar) explicitly in the class definition

class SampleClass:
    foo = staticmethod(foobar)
    def printfunc(self):
        self.foo() # Prints 'FOOBAR!'

Answer 2

The approach from user2357112's comment seems to work, too:

def foobar():
        print('FOOBAR!')

def foobaz():
        print('FooBAZ!')

class SampleClass:
        def foo(self):
                foobar()

        def printfunc(self):
                self.foo()

class DerivedClass(SampleClass):
        def foo(self):
                foobaz()

sample = SampleClass()
sample.printfunc() # FOOBAR!

derived = DerivedClass()
derived.printfunc() # FooBAZ!

If return values shall make it through, you need return statements on all levels, though:

def foobar():
        print('FOOBAR!')
        return 'foo'

def foobaz():
        print('FooBAZ!')
        return 'baz'

class SampleClass:
        def foo(self):
                return foobar()

        def printfunc(self):
                return self.foo()

class DerivedClass(SampleClass):
        def foo(self):
                return foobaz()

sample = SampleClass()
s = sample.printfunc() # FOOBAR!
print(s) # foo

derived = DerivedClass()
d = derived.printfunc() # FooBAZ!
print(d) # baz