Any method to denote object assignment?

问题

I've been studying magic methods in Python, and have been wondering if there's a way to outline the specific action of:

a = MyClass(*params).method()

versus:

MyClass(*params).method()

In the sense that, perhaps, I may want to return a list that has been split on the '\n' character, versus dumping the raw list into the variable a that keeps the '\n' intact.

Is there a way to ask Python if its next action is about to return a value to a variable, and change action, if that's the case? I was thinking:

class MyClass(object):
    def __init__(params):
        self.end = self.method(*params)

    def __asgn__(self):
        return self.method(*params).split('\n')

    def __str__(self):
        """this is the fallback if __asgn__ is not called"""
        return self.method(*params)

回答1:

No. You cannot change what happens when you assign to a bare name.

You can change what happens if the assignment target on the left hand side is an attribute or item of an object. You can override a[blah] = ... with __setitem__ and a.blah = ... with __setattr__ (although you can only hook into these on a, not on the object being assigned). But you can't override or in any way influence a = ....

Note that having the right-hand side change based on what is "going to happen" would be even stranger, and very bad. That would mean that

someFunc(MyClass().method())

could be different than

a = MyClass().method()
someFunc(a)

In Python names are just labels attached to objects. Objects don't get to know what labels are attached to them, and that's a good thing. You might assign the result a computation to an intermediate variable just to make subsequent lines more readable, and you don't want that assignment to change the result of that computation.

回答2:

There should be no difference between calling MyClass(*params).method() directly and assigning it to a variable. What you may be seeing here is your interpreter automatically printing return results, which is why it appears to be split while the variable value contains EOL markers.

There is no way to override default assignment to a variable. However, by using an object, you can easily provide your own hooks:

class Assigner(object):
    def __init__(self, assignment_callback):
        self.assignment = assignment_callback

    def __setattr__(self, key, value):
        if hasattr(self, 'assignment'):
            value = self.assignment(value)
        super(Assigner, self).__setattr__( key, value )       

def uppercase(value):
    # example function to perform on each attribute assignment
    return value.upper()

Then in your code, rather than assigning to a variable directly you assign to attributes on your object:

>>> my = Assigner(uppercase)
>>> my.a = 'foo'
>>> print my.a
FOO

来源：https://stackoverflow.com/questions/12274157/any-method-to-denote-object-assignment

标签

python

oop

magic-methods