How to compare two oracle sql object type with common base super class - sql

Question

Example:
Figure_t base class (super class)
sphere_t under figure_t
pyramid_t under figure_t
both has volume.
How to do objects comparison using map or order function?
What I am doing is using the map member function in the super class for comparing using the volume. I have tried with/without override of the map function in the subclass but still no luck. I can compare if I create the same object twice but not if I create different ones.
In example below I paste just the sphere since it is almost the same for both sphere and pyramid.
This is my super class:
CREATE OR REPLACE TYPE figure_t AS OBJECT (
v_volume NUMBER,
v_area NUMBER,
MAP MEMBER FUNCTION compare RETURN NUMBER, PRAGMA restrict_references ( compare, wnds, trust )
);
/
CREATE OR REPLACE TYPE BODY figure_t AS
MAP MEMBER FUNCTION compare RETURN NUMBER IS
BEGIN
RETURN v_volume;
END;
END;
/
ALTER TYPE figure_t NOT FINAL
CASCADE;
/
Then, this is my subtype:
CREATE OR REPLACE TYPE sphere_t UNDER figure_t (
v_radio NUMBER,
CONSTRUCTOR FUNCTION sphere_t (
radio NUMBER
) RETURN SELF AS RESULT,
MEMBER FUNCTION get_volume RETURN NUMBER,
MEMBER FUNCTION get_area RETURN NUMBER,
OVERRIDING MAP MEMBER FUNCTION compare RETURN NUMBER
);
/
CREATE OR REPLACE TYPE BODY sphere_t AS
CONSTRUCTOR FUNCTION sphere_t (
radio NUMBER
) RETURN SELF AS RESULT IS
BEGIN
self.v_radio := radio;
self.v_volume := ( 4 / 3 ) * 3.141592654 * power(radio, 3);
self.v_area := 4 * 3.141592654 * power(radio, 2);
return;
END;
MEMBER FUNCTION get_volume RETURN NUMBER IS
BEGIN
RETURN v_volume;
END;
MEMBER FUNCTION get_area RETURN NUMBER IS
BEGIN
RETURN v_area;
END;
OVERRIDING
MAP MEMBER FUNCTION compare RETURN NUMBER IS
BEGIN
RETURN self.v_volume;
END;
END;
/
For doing the comparison it looks like:
DECLARE
sphere_v sphere_t;
pyramid_v pyramid_t;
BEGIN
pyramid_v := pyramid_t(120, 90, 30);
sphere_v := sphere_t(10);
IF ( sphere_v != pyramid_v ) THEN
dbms_output.put_line('NOT EQUAL');
END IF;
END;
There should be a way for this comparison since figures have a super class in common.

Answer 1

There should be a way for this comparison since figures have a super
class in common
Am not very sure as what you wanted to achieve here. Also what is definition of pyramid_t which you are using in your comparison.
IF ( sphere_v != pyramid_v ) THEN
The above condition looks dicey to me as this will always be the case.
When you do sphere_v := sphere_t(10); means you try to get all the return of the sphere_t to sphere_v.
So it would be good if you could compare the volume and area of the Sphere and pyramid separately. See below demo how you could take these value:
DECLARE
sphere_v sphere_t;
-- pyramid_v pyramid_t;
BEGIN
sphere_v := sphere_t(10);
dbms_output.put_line('Input Radio -->'||sphere_v.v_radio);
dbms_output.put_line('Volume of Sphere-->'||sphere_v.v_volume);
dbms_output.put_line('Area Of Sphere -->'||sphere_v.v_area);
--Similarly you can take the values of `volume` and `area`
--of pyramid and get it compared with that of Sphere.
-- pyramid_v := pyramid_t(120, 90, 30);
-- dbms_output.put_line('Input Radio Pyramid -->'||pyramid_v.v_radio);
-- dbms_output.put_line('Volume of Pyramid -->'||pyramid_v.v_volume);
-- dbms_output.put_line('Area Of Pyramid -->'||pyramid_v.v_area);
-- If sphere_v.v_volume = pyramid_v.v_volume then
-- dbms_output.put_line('Equal');
-- Else
-- dbms_output.put_line('Not Equal');
END;
Assumption: pyramid_t also have the same Object Body definition having volume and area calculation.

Answer 2

There should be a way for this comparison since figures have a super class in common.
There is a way, it's just not obvious.
When the types are exactly the same e.g. two instances of the same subtype we can invoke the map function implicitly. So we can compare two spheres like this:
IF ( sphere_1 != sphere_2 ) THEN ...
However, to compare two different subtypes we need to invoke the parent map function, and to make this happen we must reference it explicitly:
IF ( sphere_v.compare() != pyramid_v.compare() ) THEN ...
Yes, this is clunky. But Oracle is an RDBMS not an ORDBMS (whatever they claimed back in the version 8.0 days).