Python Class Objects Attribute Referencing

Question

I have two classes. a and b.

In one of class a's methods, I created an object of class b. One of class b attributes takes a function. So say I gave it a random function but does this function of class b have access to class a's attribute? even though I didn't pass it in directly as a parameter?

class b:
    def __init__(self):
        self.attribute_function = None

class a:
    def __init__(self):
        self.temp = 10
        self.counter = 0

    def temp(self):
        obj = b()
        obj.attribute_function = lambda self: self.counter < self.temp
        return obj.attribute_function()


if __name__ == "__main__":
    #pass
    obj = a()
    print obj.temp()

In the above example, I tried to provide a really basic example, but if you run it, it doesn't work...

Revised Code, class a should look like this:

class a:
    def __init__(self):
        self.temp = 10
        self.counter = 0

    def temp(self):
        obj = b()
        obj.attribute_function = lambda args: self.counter < self.temp
        return obj.attribute_function(1) # i added this 1 to fill in arg

warvariuc · Accepted Answer · 2014-03-26T04:40:51.207

This works:

class b:
    def __init__(self):
        self.attribute_function = None

class a:
    def __init__(self):
        self._temp = 10
        self.counter = 0

    def temp(self):
        obj = b()
        obj.attribute_function = lambda self=self: self.counter < self._temp
        return obj.attribute_function()


if __name__ == "__main__":
    obj = a()
    print obj.temp()

On problem you had is self.temp = 10 which shadowed your method temp().

Another problem: lambda self: self.counter < self._temp. Your lambda function was expecting an argument. But omitting self is not a good idea lambda : self.counter < self._temp, because if you call obj.attribute_function() somewhere where self is not available or has changed - it will not find self or use another self. self=self fixes that.

But generally such magic is an anti-pattern. Tell us what are your trying to achieve, and there should be a better way to do what you want. Otherwise this kind of code will ensure many headaches.

I think this is a better solution (called strategy pattern):

class B:
    def __init__(self, a):
        self.a = a
    def temp(self):
        return self.a.temp()

class A:
    def __init__(self):
        self._temp = 10
        self.counter = 0

    def temp(self):
        return self.counter < self._temp


if __name__ == "__main__":
    obj = B(A())
    print obj.temp()

I apologize for the bad sample code. Im working on a rather complicated project and explaining it over SO is tough. But to correct my error, the `attribute_function`'s `self` parameter in my original code is another parameter for something entirely different. the `self` reference within the `attribute_function` is entirely from the creator class's attribute. — user1234440, Mar 26 '14 at 04:50
Please see my above revised code; with you help I get it. But just to confirm, referencing the attributes of that class `a` without implicitly passing in the parameter works correct? Is this safe? — user1234440, Mar 26 '14 at 04:54
> But just to confirm, referencing the attributes of that class a without implicitly passing in the parameter works correct? Is this safe?< It should work, and this is called `closure`. Think twice if you need it. The only think I would do: `obj.attribute_function = lambda args, self=self: self.counter < self.temp` to ensure that the correct `self` is used when your lambda is called. See [this question](http://stackoverflow.com/questions/2295290/what-do-lambda-function-closures-capture-in-python) — warvariuc, Mar 26 '14 at 06:38

score 0 · Answer 2 · answered Mar 26 '14 at 04:35

0

Your example does not work because you have a name collision at temp

You have assigned temp to be both a method:

def temp(self):

and an attribute:

self.temp = 10

answered Mar 26 '14 at 04:35

qwwqwwq

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Python Class Objects Attribute Referencing

2 Answers2