Python的类继承

Question

# Defining a Base class to be shared among many other classes later:

class Base(dict):
    """Base is the base class from which all the class will derrive.
    """
    name = 'name'    
    def __init__( self):
        """Initialise Base Class
        """
        dict.__init__(self)
        self[Base.name] = ""

# I create an instance of the Base class:

my_base_instance = Base()

# Since a Base class inherited from a build in 'dict' the instance of the class is a dictionary. I can print it out with:

print my_base_instance   Results to: {'name': ''}


# Now I am defining a Project class which should inherit from an instance of Base class:

class Project(object):
    def __init__(self):
        print "OK"
        self['id'] = ''

# Trying to create an instance of Project class and getting the error:

project_class = Project(base_class)

TypeError: __init__() takes exactly 1 argument (2 given)

Answer 1

当实例化一个类时，不需要传递base_class 。 这是按定义完成的。 __init__接受1个参数，它是self ，并且是自动的。 你只需要打电话

project_class = Project()

Answer 2

为了使Project从Base继承，您不应从Object继承它，而应从Base继承它，即class Project(Base) 。 您会得到TypeError: init() takes exactly 1 argument (2 given)实例化Project类时， TypeError: init() takes exactly 1 argument (2 given)错误，因为构造函数仅接受1个参数（ self ），并且还传递了base_class 。 'self'由python隐式传递。

Answer 3

您的代码中有两个错误：

1）类继承

class Project(Base):   # you should inherit from Base here...
    def __init__(self):
        print "OK"
        self['id'] = ''

2）实例定义（您的__init__不需要任何显式参数，并且当然不需要祖先类）

project_class = Project() # ...and not here since this is an instance, not a Class