How to call a property of the base class if this property is being overwritten in the derived class?
Solution 1:
You might think you could call the base class function which is called by property:
class FooBar(Foo):
@property
def bar(self):
# return the same value
# as in the base class
return Foo.bar(self)
Though this is the most obvious thing to try I think - it does not work because bar is a property, not a callable.
But a property is just an object, with a getter method to find the corresponding attribute:
class FooBar(Foo):
@property
def bar(self):
# return the same value
# as in the base class
return Foo.bar.fget(self)
Solution 2:
super()
should do the trick:
return super().bar
In Python 2.x you need to use the more verbose syntax:
return super(FooBar, self).bar
Solution 3:
There is an alternative using super
that does not require to explicitly reference the base class name.
Base class A:
class A(object):
def __init__(self):
self._prop = None
@property
def prop(self):
return self._prop
@prop.setter
def prop(self, value):
self._prop = value
class B(A):
# we want to extend prop here
pass
In B, accessing the property getter of the parent class A:
As others have already answered, it's:
super(B, self).prop
Or in Python 3:
super().prop
This returns the value returned by the getter of the property, not the getter itself but it's sufficient to extend the getter.
In B, accessing the property setter of the parent class A:
The best recommendation I've seen so far is the following:
A.prop.fset(self, value)
I believe this one is better:
super(B, self.__class__).prop.fset(self, value)
In this example both options are equivalent but using super has the advantage of being independent from the base classes of B
. If B
were to inherit from a C
class also extending the property, you would not have to update B
's code.
Full code of B extending A's property:
class B(A):
@property
def prop(self):
value = super(B, self).prop
# do something with / modify value here
return value
@prop.setter
def prop(self, value):
# do something with / modify value here
super(B, self.__class__).prop.fset(self, value)
One caveat:
Unless your property doesn't have a setter, you have to define both the setter and the getter in B
even if you only change the behaviour of one of them.
Solution 4:
try
@property
def bar:
return super(FooBar, self).bar
Although I'm not sure if python supports calling the base class property. A property is actually a callable object which is set up with the function specified and then replaces that name in the class. This could easily mean that there is no super function available.
You could always switch your syntax to use the property() function though:
class Foo(object):
def _getbar(self):
return 5
def _setbar(self, a):
print a
bar = property(_getbar, _setbar)
class FooBar(Foo):
def _getbar(self):
# return the same value
# as in the base class
return super(FooBar, self)._getbar()
def bar(self, c):
super(FooBar, self)._setbar(c)
print "Something else"
bar = property(_getbar, _setbar)
fb = FooBar()
fb.bar = 7
Solution 5:
Some small improvements to Maxime's answer:
- Using
__class__
to avoid writingB
. Note thatself.__class__
is the runtime type ofself
, but__class__
withoutself
is the name of the enclosing class definition.super()
is a shorthand forsuper(__class__, self)
. - Using
__set__
instead offset
. The latter is specific toproperty
s, but the former applies to all property-like objects (descriptors).
class B(A):
@property
def prop(self):
value = super().prop
# do something with / modify value here
return value
@prop.setter
def prop(self, value):
# do something with / modify value here
super(__class__, self.__class__).prop.__set__(self, value)