On 07/11/2011 05:54 PM, Anthony Kong wrote:
> Hi, all,
>
> This question is in the same context of my two earlier questions. This
> question was raised by some python beginners, and I would like to check
> with the list to ensure I provide a correct answer.
>
> Here is a code snippet I used to demonstrate the keyword *property*:
>
>
> class A(object):
>
> def __init__(self):
> self.__not_here = 1
>
> def __get_not_here(self):
> return self.__not_here
>
> def __set_not_here(self, v):
> print "I am called"
> self.__not_here = v
>
> not_here = property(lambda self: self.__get_not_here(), lambda self,
> v: self.__set_not_here(v))
> # not_here = property(lambda self: self.__not_here, lambda self, v:
> self.__not_here = v)
>
> So the question: is it possible to use lambda expression at all for the
> setter? (As in the last, commented-out line)
>
> Python interpreter will throw an exception right there if I use the last
> line ('SyntaxError: lambda cannot contain assignment'). I'd use pass a
> setter method anyway.
>
> What is your preferred solution?
No, a lambda can only contain an expression, not a statement. This is
not C, assignments are not expressions.
As to what I would do:
There's really no need to use lambdas at all here:
class A(object):
def __init__(self):
self.not_here = 1
def __get_not_here(self):
return self.__not_here
def __set_not_here(self, val):
self.__not_here = val
not_here = property(__get_not_here, __set_not_here)
My favourite way to create properties is of course with decorators:
class A(object):
def __init__(self):
self.not_here = 1
@property
def not_here(self):
return self.__not_here
@not_here.setter
def not_here(self, val):
self.__not_here = val
--
http://mail.python.org/mailman/listinfo/python-list
