在Python中创建简单的对象层次结构时，我希望能够从派生类调用父类的方法。在Perl和Java中，对此有一个关键字(super)。在Perl中，我可以这样做:
package Foo;
sub frotz {
return "Bamf";
}
package Bar;
@ISA = qw(Foo);
sub frotz {
my $str = SUPER::frotz();
return uc($str);
}
在Python中，似乎我必须显式地从子类中命名父类。 在上面的例子中，我必须做一些类似于Foo::frotz()的事情。
这似乎是不正确的，因为这种行为使得很难建立深层的等级制度。如果孩子们需要知道哪个类定义了一个继承的方法，那么就会产生各种各样的信息麻烦。
这是python的实际限制，还是我理解上的差距，或者两者兼而有之?
当前回答
下面是一个使用super()的例子:
#New-style classes inherit from object, or from another new-style class
class Dog(object):
name = ''
moves = []
def __init__(self, name):
self.name = name
def moves_setup(self):
self.moves.append('walk')
self.moves.append('run')
def get_moves(self):
return self.moves
class Superdog(Dog):
#Let's try to append new fly ability to our Superdog
def moves_setup(self):
#Set default moves by calling method of parent class
super(Superdog, self).moves_setup()
self.moves.append('fly')
dog = Superdog('Freddy')
print dog.name # Freddy
dog.moves_setup()
print dog.get_moves() # ['walk', 'run', 'fly'].
#As you can see our Superdog has all moves defined in the base Dog class
其他回答
下面是一个使用super()的例子:
#New-style classes inherit from object, or from another new-style class
class Dog(object):
name = ''
moves = []
def __init__(self, name):
self.name = name
def moves_setup(self):
self.moves.append('walk')
self.moves.append('run')
def get_moves(self):
return self.moves
class Superdog(Dog):
#Let's try to append new fly ability to our Superdog
def moves_setup(self):
#Set default moves by calling method of parent class
super(Superdog, self).moves_setup()
self.moves.append('fly')
dog = Superdog('Freddy')
print dog.name # Freddy
dog.moves_setup()
print dog.get_moves() # ['walk', 'run', 'fly'].
#As you can see our Superdog has all moves defined in the base Dog class
class department:
campus_name="attock"
def printer(self):
print(self.campus_name)
class CS_dept(department):
def overr_CS(self):
department.printer(self)
print("i am child class1")
c=CS_dept()
c.overr_CS()
Python中也有一个super()。它有点不稳定，因为Python有新旧风格的类，但它非常常用，例如在构造函数中:
class Foo(Bar):
def __init__(self):
super(Foo, self).__init__()
self.baz = 5
在python中也有一个super()。
如何从子类方法调用父类方法的示例
class Dog(object):
name = ''
moves = []
def __init__(self, name):
self.name = name
def moves_setup(self,x):
self.moves.append('walk')
self.moves.append('run')
self.moves.append(x)
def get_moves(self):
return self.moves
class Superdog(Dog):
#Let's try to append new fly ability to our Superdog
def moves_setup(self):
#Set default moves by calling method of parent class
super().moves_setup("hello world")
self.moves.append('fly')
dog = Superdog('Freddy')
print (dog.name)
dog.moves_setup()
print (dog.get_moves())
这个例子类似于上面解释的例子。但是有一个区别，super没有传递任何参数给它。以上代码在python 3.4版本中可执行。
如果你不知道你可能会得到多少争论，并且想把它们都传递给孩子:
class Foo(bar)
def baz(self, arg, *args, **kwargs):
# ... Do your thing
return super(Foo, self).baz(arg, *args, **kwargs)
(来自:Python -覆盖__init__的最干净的方法，其中可选的kwarg必须在super()调用后使用?)
