参数如何通过 __getattr__ 传递给函数 [英] How are arguments passed to a function through __getattr__

问题描述

考虑以下代码示例(python 2.7):

class 父类:def __init__(self, child):self.child = 孩子def __getattr__(self, attr):print("调用__getattr__:"+attr)如果 hasattr(self.child, attr):返回 getattr(self.child, attr)别的:引发 AttributeError(attr)班级儿童:def make_statement(self, age=10):print("我是一个Child with age "+str(age))孩子 = 孩子()人 = 父母(孩子)Kid.make_statement(5)person.make_statement(20)

可以看出，函数调用person.make_statement(20)通过Parent的调用了Child.make_statement函数__getattr__ 函数.在 __getattr__ 函数中，我可以在调用子实例中的相应函数之前打印出属性.到此为止很清楚.

但是调用 person.make_statement(20) 的参数是如何通过 __getattr__ 传递的?如何在 __getattr__ 函数中打印出数字20"?

解决方案

您没有在 __getattr__ 函数中打印 20.该函数在 Child 实例上找到 make_statement attribute 并返回它.碰巧，该属性是一个方法，因此它是可调用的.Python因此调用返回的方法，然后那个方法打印20.

如果你删除了 () 调用，它仍然可以工作；我们可以存储该方法并单独调用它以打印20:

<预><代码>>>>person.make_statement调用 __getattr__:make_statement<0x10db5ed88处<__main__.Child实例的绑定方法Child.make_statement>>>>>ms = person.make_statement调用 __getattr__:make_statement>>>多发性硬化症()我是 10 岁的 Child 的一个实例

如果您必须查看参数，则必须返回一个包装函数:

def __getattr__(self, attr):print("调用__getattr__:"+attr)如果 hasattr(self.child, attr):def 包装器(*args, **kw):print('用 %r 和 %r' % (args, kw))返回 getattr(self.child, attr)(*args, **kw)返回包装器引发 AttributeError(attr)

现在结果:

<预><代码>>>>person.make_statement(20)调用 __getattr__:make_statement使用 (20,) 和 {} 调用我是 20 岁的 Child 的一个实例

Consider the following code example (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)

it can be shown, that the function call person.make_statement(20) calls the Child.make_statement function through the Parent's __getattr__ function. In the __getattr__ function I can print out the attribute, before the corresponding function in the child instance is called. So far so clear.

But how is the argument of the call person.make_statement(20) passed through __getattr__? How am I able to print out the number '20' in my __getattr__ function?

解决方案

You are not printing 20 in your __getattr__ function. The function finds the make_statement attribute on the Child instance and returns that. As it happens, that attribute is a method, so it is callable. Python thus calls the returned method, and that method then prints 20.

If you were to remove the () call, it would still work; we can store the method and call it separately to get 20 printed:

>>> 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

If you have to see the arguments, you'd have to return a wrapper function instead:

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)

This now results in:

>>> person.make_statement(20)
Calling __getattr__: make_statement
called with (20,) and {}
I am an instance of Child with age 20

这篇关于参数如何通过 __getattr__ 传递给函数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！