先上一个比较简单的单继承语法。在python3中,基类的构造函数不会被自动调用,需要手动调用,同样的方法也是这样,需要手动调用。可以使用类名称+init方法,也可以使用super语法进行调用。在下面这个例子中,子类继承了基类的方法和字段。字段会在基类中初始化。
class BaseClass:
def __init__(self):
self.name = 'BaseClass'
print('BaseCalss: Constructor called')
def getname(self):
print('BaseCalss: self name equals ' + self.name)
class DerivedClass(BaseClass):
def __init__(self):
super().__init__()
print('DerivedClass: Constructor called')
if __name__ == '__main__':
class1 = BaseClass()
class1.getname()
class2 = DerivedClass()
class2.getname()
运行结果:
BaseCalss: Constructor called BaseCalss: self name equals BaseClass BaseCalss: Constructor called DerivedClass: Constructor called BaseCalss: self name equals BaseClass
子类也可以overwrite父类的方法,那么父类的方法就不会被调用,除非手动调用:
class BaseClass:
def __init__(self):
self.name = 'BaseClass'
print('BaseCalss: Constructor called')
def getname(self):
print('BaseCalss: self name equals ' + self.name)
class DerivedClass(BaseClass):
def __init__(self):
super().__init__()
print('DerivedClass: Constructor called')
def getname(self):
print('self.name init value is ' + self.name)
self.name = 'DerivedClass'
print('DerivedClass: self name equals ' + self.name)
if __name__ == '__main__':
class1 = BaseClass()
class1.getname()
class2 = DerivedClass()
class2.getname()
运行结果:
BaseCalss: Constructor called
BaseCalss: self name equals BaseClass
BaseCalss: Constructor called
DerivedClass: Constructor called
self.name init value is BaseClass
DerivedClass: self name equals DerivedClass
python不仅仅支持单继承,还支持多继承,字段和方法都可以被继承。在多继承super()只能代表继承的第一个父类,所以您在子类的构造函数中,不能单独使用super().__init__(), 那只是表示调用其中一个基类的构造函数。所以用super就不是那么好用了。还是要用会原来的类名+init方法来调用。
class BaseClass1:
def __init__(self):
self.name1 = 'BaseClass1_Name1'
self.name = 'BaseClass1_Name'
print('BaseCalss1: Constructor called')
def getname1(self):
print('BaseCalss1: self name1 equals ' + self.name1)
def getname(self):
print('BaseCalss1: getname called, name equal ' + self.name)
class BaseClass2:
def __init__(self):
self.name2 = 'BaseClass2_Name2'
self.name = 'BaseClass2_Name'
print('BaseClass2: Constructor called')
def getname2(self):
print('BaseClass2: self name2 equals ' + self.name2)
def getname(self):
print('BaseCalss2: getname called, name equal ' + self.name)
class DerivedClass2(BaseClass1, BaseClass2):
def __init__(self):
BaseClass1.__init__(self)
BaseClass2.__init__(self)
print('DerivedClass: Constructor called')
if __name__ == '__main__':
class1 = BaseClass1()
class1.getname1()
class2 = BaseClass2()
class2.getname2()
class3 = DerivedClass2()
class3.getname1()
class3.getname2()
class3.getname()
运行结果:
BaseCalss1: Constructor called
BaseCalss1: self name1 equals BaseClass1_Name1
BaseClass2: Constructor called
BaseClass2: self name2 equals BaseClass2_Name2
BaseCalss1: Constructor called
BaseClass2: Constructor called
DerivedClass: Constructor called
BaseCalss1: self name1 equals BaseClass1_Name1
BaseClass2: self name2 equals BaseClass2_Name2
BaseCalss1: getname called, name equal BaseClass2_Name
我们可以看到,当两个基类有方法重名的时候,python3会按照您继承类的从左到右的顺序查找您调用的方法DerivedClass2(BaseClass1, BaseClass2)。在这个例子中,是先找BaseClass1,然后再找BaseClass2。
如果您的代码需要多层继承的话,可以参开多重继承的 Diamond Problem 问题。
class A1:
def foo1(self):
print("Call A1's foo1")
class A2:
def foo1(self):
print("Call A2's foo1")
def foo2(self):
print("Call A2's foo2")
class B1(A1,A2):
pass
class B2(A1,A2):
def foo2(self):
print("Call B2's foo2")
class C(B1,B2):
pass
if __name__ == '__main__':
class1 = C()
class1.foo1()
class1.foo2()
运行结果:
Call A1's foo1
Call B2's foo2
所以对于python3 的多层继承来说,因为都是新式类,总是从左到右,广度优先的方式进行。
本例子代码可以从这里下载。