python 中接口的实现

实际上,由于python是动态语言,支持多继承,因此接口在语言层面,是没有的东东。

然后,在架构设计上,我们又需要这么一个东西,来规范开发人员的行为。

因此就有了zope.interface的诞生。

定义接口

继承 zope.interface.Interface即可,如下:

import zope.interface
class IFoo(zope.interface.Interface):
	"""Foo blah blah"""
	x = zope.interface.Attribute("""x blah blah""")
	def bar(q, r=None):
		"""bar blah blah"""

1. 接口不是类

>>> type(IFoo)
<class 'zope.interface.interface.InterfaceClass'>
>>> class A(object):
	pass

>>> type(A)
<type 'type'>

但能正常访问:__dict__ / __name__ / __module__ / __doc__ 等专有属性

 

2. 行为有点类似于字典

>>> list(IFoo)
['x', 'bar']
>>> IFoo.get('bar')
<zope.interface.interface.Method object at 0x0000000002B7B710>
>>> callable(IFoo.get('bar'))
True

3. 接口中的方法能自动获取签名

>>> IFoo['bar'].getSignatureString()
'(q, r=None)'

 

实现接口

在声明接口之前,有两个术语要说明下:

  provide(提供接口)

         对象实例(object)提供接口,接口详细描述/规范了对象的行为。

  implement(实现)

         类(可以抽象成工厂类)实现(implement)接口。

         类一般不会提供(provide)接口。只有对象提供接口。【备注:此处属于术语概念,了解即可】

1. 实现接口:

 a). 用类实现

>>> class Foo:
	zope.interface.implements(IFoo)

有点诧异:没有报错!对象实例呢?

>>> f = Foo()

也没有报错!然后下面却报错了:

>>> IFoo['bar'](1)

Traceback (most recent call last):
  File "<pyshell#85>", line 1, in <module>
    IFoo['bar'](1)
  File "C:\Python27\lib\site-packages\zope\interface\interface.py", line 616, in __call__
    raise BrokenImplementation(self.interface, self.__name__)
BrokenImplementation: An object has failed to implement interface <InterfaceClass __main__.IFoo>

        The bar attribute was not provided.

问题:该如何在子类的实例中使用接口里面定义的方法?

 

如何规范开发人员的行为?

答案是:通过测试来发现对象或者类是否实现了定义的接口。通过代码测试来做。通过测试控制开发,似乎有点马后炮。

主要使用 zope.interface.verify 里面的 verifyObject 和 verifyClass方法。可以参考这里

>>> class IFoo(Interface):
	x = Attribute("The X attribute")
	def bar(q,r=None):
		"""bar of interface"""

		
>>> class Foo:
	implements(IFoo)

	
>>> from zope.interface.verify import verifyObject, verifyClass
>>> verifyObject(IFoo, Foo())

Traceback (most recent call last):
  File "<pyshell#13>", line 1, in <module>
    verifyObject(IFoo, Foo())
  File "C:\Python27\lib\site-packages\zope\interface\verify.py", line 105, in verifyObject
    return _verify(iface, candidate, tentative, vtype='o')
  File "C:\Python27\lib\site-packages\zope\interface\verify.py", line 62, in _verify
    raise BrokenImplementation(iface, name)
BrokenImplementation: An object has failed to implement interface <InterfaceClass __main__.IFoo>

        The x attribute was not provided.
        
>>> verifyClass(IFoo, Foo)

Traceback (most recent call last):
  File "<pyshell#14>", line 1, in <module>
    verifyClass(IFoo, Foo)
  File "C:\Python27\lib\site-packages\zope\interface\verify.py", line 102, in verifyClass
    return _verify(iface, candidate, tentative, vtype='c')
  File "C:\Python27\lib\site-packages\zope\interface\verify.py", line 62, in _verify
    raise BrokenImplementation(iface, name)
BrokenImplementation: An object has failed to implement interface <InterfaceClass __main__.IFoo>

        The bar attribute was not provided.

  

 

 

b) 用实例变量实现:

>>> b= Foo()
>>> b = zope.interface.implementer(IFoo)(b)
>>> list(zope.interface.providedBy(b))
[<InterfaceClass __main__.IFoo>]

 

 

在python 2.6后,可以直接用修饰符,上面代码的等同代码如下:

>>> from zope.interface import implementer
>>> @implementer(IFoo)
class Foo:
	pass

>>> list(zope.interface.implementedBy(Foo))
[<InterfaceClass __main__.IFoo>]

  

 

2. 查看类实现了哪些接口

>>> list(zope.interface.implementedBy(Foo))
[<InterfaceClass __main__.IFoo>]

  测试某接口是否由某类实现:

>>> IFoo.implementedBy(Foo)
True

3. 查看对象实例实现了哪些接口

>>> list(zope.interface.providedBy(f))
[<InterfaceClass __main__.IFoo>]

  测试某接口是否由某对象提供

>>> IFoo.providedBy(f)
True

  

 接口的继承

与python的类继承相似。又有所不同,接口继承不是深度优先,如下:

>>> import zope.interface
>>> class IBase(zope.interface.Interface):
	def foo():
		"""base foo doc"""

		
>>> class IBase1(IBase):
	pass

>>> class IBase2(IBase):
	def foo():
		"""base2 foo doc"""

		
>>> class ISub(IBase1, IBase2):
	pass

>>> ISub['foo'].__doc__
'base2 foo doc'

  

类的继承:

>>> class Base:
	def foo(self):
		print "base"

		
>>> class Base1(Base):
	pass

>>> class Base2(Base):
	def foo(self):
		print "base2"

		
>>> class Sub(Base1,Base2):
	pass

>>> Sub().foo()
base

  

 

posted @ 2014-11-14 16:23  tommy.yu  阅读(2188)  评论(0编辑  收藏  举报