I have a base class in Ruby that has a class method which it has inherited. I would like to call this method in the base class, but pass it an option which is specified by the derived class, like so:
class Base < SuperDuperClass
super_duper_class_method :option => my_option_value
def self.my_option_value
raise "Method my_option_value must be overridden by all subclasses"
end
end
class Derived < Base
def self.my_option_value
"My Derived Option Value"
end
end
However, this isn't working. I believe it's because the top-level code in the base class is executed before the top-level code in the derived class, so the derived method is not defined when super_duper_class_method is called. I'd rather not have to call super_duper_class_method in all the derived classes, but just specify the option instead.
Any ideas?
You weren't specific when you said it wasn't working. I believe you will get an exception raised when the Base class is defined, (rather than when my_option_value is called).
I suggest that you pass the method name as a symbol, and in the SuperDuperClass, call send on self to call the derived method.
class SuperDuperMethod
def self.some_code
self.class.__send__(options[:option])
end
...
Just to close this off, my solution is to call super_duper_class_method
in the subclasses. This is what I was trying to avoid, but in this situation, I don't think it's possible to get what I was going for without some pretty wild metaprogramming.
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