考虑下面的代码示例(
python 2.7):
- class Parent:
- def __init__(self,child):
- self.child = child
- def __getattr__(self,attr):
- print("Calling __getattr__: "+attr)
- if hasattr(self.child,attr):
- return getattr(self.child,attr)
- else:
- raise AttributeError(attr)
- class Child:
- def make_statement(self,age=10):
- print("I am an instance of Child with age "+str(age))
- kid = Child()
- person = Parent(kid)
- kid.make_statement(5)
- person.make_statement(20)
可以显示,函数调用person.make_statement(20)通过Parent的__getattr__函数调用Child.make_statement函数.在__getattr__函数中,我可以在子实例的相应函数被调用之前打印出属性.到目前为然这么清楚
但是调用person.make_statement(20)的参数怎么通过__getattr__?我可以在__getattr__函数中打印出数字“20”吗?
解决方法
你不是在__getattr__函数中打印20.该函数在Child实例上找到make_statement属性并返回.发生这种情况,该属性是一种方法,因此它是可调用的. Python因此调用返回的方法,然后打印20.
如果你要删除()调用,它仍然可以工作;我们可以存储方法并单独打电话给20打印:
- >>> person.make_statement
- Calling __getattr__: make_statement
- <bound method Child.make_statement of <__main__.Child instance at 0x10db5ed88>>
- >>> ms = person.make_statement
- Calling __getattr__: make_statement
- >>> ms()
- I am an instance of Child with age 10
如果你必须看到参数,你必须返回一个包装函数:
- def __getattr__(self,attr):
- print("Calling __getattr__: "+attr)
- if hasattr(self.child,attr):
- def wrapper(*args,**kw):
- print('called with %r and %r' % (args,kw))
- return getattr(self.child,attr)(*args,**kw)
- return wrapper
- raise AttributeError(attr)
现在这样做:
- >>> person.make_statement(20)
- Calling __getattr__: make_statement
- called with (20,) and {}
- I am an instance of Child with age 20