python 中的方法与类相关联。 与实例直接关联的函数意味着什么？

Question

I know that in python, methods are associated the class and not the instance but they accept the instance as their first argument so as to access it. but, one can still define a function and tie it to an instance directly. What does this imply? and can this function somehow access the instance?

class TestClass:
    def __init__(self, a):
       self.a = a

    def method1(self):
        return self.a

my_obj = TestClass(12)

print(my_obj.method1()) // 12

Python also allows this:

class TestClass:
    def __init__(self, a):
       self.a = a
my_obj = TestClass(12)

def method1_defined_outside(self):
    return self.a

TestClass.method1 = method1_defined_outside

print(my_obj.method1()) // 12

The two snippets above mean the same thing. but python also allows this:

class TestCase:
    def __init__(self, a):
       self.a = a
my_obj = TestCase(12)

def method1_defined_outside(self):
    return self.a

my_obj.method1 = method1_defined_outside

print(my_obj.method1()) // this throws an error stating the method1 is missing one postional argument(s), so clearly self does not automatically equate to the object

difference between the second and the third snippet is that the method defined outside is tied to the instance in latter, and to the class in the former.

Answer 1

When you access a method through an instance, but the method lives on the class, accessing the method actually creates a method wrapper. This wrapper knows what instance the method was called on, and passes it through to the actual function as the first argument (named, by convention, self ). The mechanism behind this is the descriptor protocol and it's how properties work, too.

When you put the method directly on the instance, none of that applies and it's just a function that you access by way of the instance. It doesn't get self .

Now, you can create a method wrapper by manually invoking the descriptor, and set the resulting bound method on the instance.

my_obj.method1 = method1_defined_outside.__get__(my_obj)

Now my_obj.method1() passes my_obj as self .

The situations in which you'd want to set a method on an instance like this are fairly rare, though.

Answer 2

One way you could force sending the instance as self is by using the functools.partial wrapper as:

from functools import partial

class A:
    def __init__(self):
        self.x = 10

a1 = A()

def do_external(self):
    print("EXTERNAL", self.x)

a1.doext = partial(do_external, a1)

a1.doext()

which will yield in the console:

EXTERNAL 10