如何可通过__getattr__传递给函数的参数 [英] How are arguments passed to a function through __getattr__

问题描述

考虑以下code示例（Python 2.7）：

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)

可以证明，该函数调用 person.make_statement（20）调用 Child.make_statement 功能通过父的 __ __ GETATTR 功能。在 __ __ GETATTR 的功能，我可以打印出属性，在孩子的实例对应的函数被调用之前。到目前为止那么明确了。

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.

但是如何调用的参数 person.make_statement（20）到 __ __ GETATTR 通过呢？我怎么能在我的 __ __ GETATTR 函数打印出数字'20'？

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?

推荐答案

您不打印 20 在 __ GETATTR __ 功能。该功能可查找 make_statement 的属性的child实例并返回。碰巧的是，这个属性是一个方法，所以它是可调用的。蟒蛇从而调用返回的方法，而是的方法，然后打印 20 。

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.

如果您要删除（）调用，它仍能正常工作;我们可以存储方法并分别调用它来获得 20 打印：

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

如果您的有无的看参数，你必须返回一个包装函数来代替：

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)

这现在将会导致：

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

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