@property setter not called in python 3

Question

I have a big issue, I'm new with @property and setter but I need to use them for a University Assignment: my problem is that the setters are always bypassed by the getters and also, when i try to pass an argument to the setter, it doesn't work:

class DumbClass(): def __init__(self, classParam): self.ciao = classParam @property def getCiao(self): return self.ciao @getCiao.setter def setCiao(self,dummy): self.ciao = dummy

Then, when i call it I have either:

pizzo = DumbClass('getter') pozza = pizzo.getCiao('setter') print(pizzo.ciao, pozza) ---------------------------------------------------------------------- TypeError Traceback (most recent call last) <ipython-input-18-78a6b4607757> in < module>() 1 pizzo = DumbClass('getter') ----> 2 pozza = pizzo.getCiao('setter') 3 print(pizzo.ciao, pozza) TypeError: 'str' object is not callable

Or, if I don't pass any arguments:

pizzo = DumbClass('getter') pozza = pizzo.getCiao print(pizzo.ciao, pozza) ---------------------------------------------------------------------- getter getter

I can I make the setter to be called?

You've defined a property called getCiao, not ciao. ciao is still just an attribute. — khelwood
– khelwood, Commented Nov 15, 2018 at 15:45
Thank you Khelwood, I knew that there was something wrong with that... — Fabio Magarelli
– Fabio Magarelli, Commented Nov 15, 2018 at 15:58

JBernardo · Accepted Answer · 2018-11-15 15:46:22Z

4

Your property name should be ciao and the real variable something like _ciao So when you do self.ciao = ... it will call the property

class DumbClass(): def __init__(self, classParam): self.ciao = classParam @property def ciao(self): return self._ciao @ciao.setter def ciao(self, dummy): self._ciao = dummy

answered Nov 15, 2018 at 15:46

JBernardo

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2 Comments

JBernardo Over a year ago

@MosesKoledoye The explaining is quite simple and in the 1st line of this answer. The property name should be the one he wants to use.

Moses Koledoye Over a year ago

Although a misnomer, they can do pizzo.getCiao = ... to set the property. OP seems to be misunderstanding how properties work, not just the naming convention which you've very well explained.

chepner · Accepted Answer · 2018-11-15 15:57:19Z

The setter is only invoked when you try to assign to the property. In the line

pozza = pizzo.getCiao('setter')

you first access the value of the property with pizzo.getCiao (which calls the getter and returns the string "getter", then attempts to call the string as a function with the argument 'setter'.

You have basially created a read-only property:

>>> pizzo = DumbClass('getter') >>> pizzo.getCiao 'getter' >>> pizzo.getCiao = 'foo' Traceback (most recent call last): File "<stdin>", line 1, in <module> AttributeError: can't set attribute

This is because you didn't use the same name when defining both the getter and setter, so your property doesn't have a setter. (Arguably, getCiao.setter should probably raise an error right away, since the name matters, but it doesn't. C'est la vie.)

So the right thing to do is use the same name for both. The usual convention is to use an private variable with the "same" name as the property to store the underlying data for simple properties like this.

class DumbClass: def __init__(self, p): self._ciao = p @property def ciao(self): return self._ciao @ciao.setter def ciao(self, v): self._ciao = v

Now you should be able to both get and set the value of the propery.

>>> d = DumbClass("hi") >>> d.ciao 'hi' >>> d.ciao = 'bye' >>> d.ciao 'bye'

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