方法分配和对象

我有一个 python 问题： 我想将一个方法分配给另一个类的对象，但在这个方法中使用它自己的属性。由于我的项目中有许多具有不同使用方法的容器（不在该示例中），因此我不想使用继承，这将迫使我为每个实例创建一个自定义类。

class container():
    def __init__(self):
        self.info = "undefiend info attribute"

    def use(self):
        print self.info


class tree():
    def __init__(self):

        # create container instance
        b = container()

        # change b's info attribute
        b.info = "b's info attribute"

        # bound method test is set as use of b and in this case unbound, i think
        b.use = self.test

        # should read b's info attribute and print it
        # should output: test: b's info attribute but test is bound in some way to the tree object
        print b.use()

    # bound method test
    def test(self):
        return "test: "+self.info


if __name__ == "__main__":
    b = tree()

非常感谢您阅读本文，也许对我有帮助！ :)

干得好。您应该知道 self.test 已经绑定，因为当您进入 __init__ 时，实例已经创建并且其方法已绑定。因此，您必须使用 im_func 成员访问未绑定的成员，并将其与 MethodType 绑定。

import types

class container():
    def __init__(self):
        self.info = "undefiend info attribute"

    def use(self):
        print self.info


class tree():
    def __init__(self):

        # create container instance
        b = container()

        # change b's info attribute
        b.info = "b's info attribute"

        # bound method test is set as use of b and in this case unbound, i think
        b.use = types.MethodType(self.test.im_func, b, b.__class__)

        # should read b's info attribute and print it
        # should output: test: b's info attribute but test is bound in some way to the tree object
        print b.use()

    # bound method test
    def test(self):
        return "test: "+self.info


if __name__ == "__main__":
    b = tree()