I read that :
whenever a collection need to be sorted, the elements must be mutually comparable.
I wrote the below code and it
It means that for any two items in the collection, you must be able to compare A to B and vice versa.
Mutually comparable means that instances of both classes must be comparable to the other.
In case of type A
and type B
:
A
must implement Comparable<T>
where T
extends B
(or is B
)B
must implement Comparable<U>
where U
extends A
(or is A
)a>b
then it must be so b<a
. And if a<b
then it must be so b>a
. And if a==b
then it must be so b==a
.So that the following code can be written:
A a = ...;
B b = ...;
int rab = a.compareTo(b); // A is comparable to B
int rba = b.compareTo(a); // B is comparable to A
// And also consistent:
if ( !(rab<0 && rba>0)
|| !(rab>0 && rba<0)
|| !(rab==0 && rba!=0))
System.out.println("Not consistent!");
If type A
and type B
are the same, then a
and b
are obviously mutually comparable.
Sorting algorithms usually require the elements to be mutually comparable so they can compare any 2 of the sortable elements to each other.
Back to your question
Your case is special. Your 2 objects obj1
and obj2
are mutually comparable according to the definition. Your obj1
and obj2
instances are mutually comparable, but your b
and c
classes are not.
However, if you would pass another instance of either type a
or type b
, all elements would be not. Because if we would select 2 instances from the collection of the same type, they would not implement Comparable
to the same type.
obj1.compareTo(obj2); // This is valid, b implements Comparable<c>
obj2.compareTo(obj1); // This is also valid, c implements Comparable<b>
obj1.compareTo(obj1); // This is NOT valid! b does not implement Comparable<b>
Your code works only by chance and it has so many mistakes that they cancel out each other resulting into working code.
ArrayList list = new ArrayList();
What is a raw type and why shouldn't we use it?Comparable<T>
, T
should be a super type of that class. For example Integer implements Comparable<Number>
would be just as fine.the elements must be mutually comparable
That means that each element from that collection must be comparable to each other and in your case that works only because you added only 2 elements (adding more would result into ClassCastException at runtime).
If you want a sortable collection containing both B
and C
, find common parent of those classes (or perhaps create an interface for it, lets call it BC
) then you would go:
class B implments BC, Comparable<BC> { ... }
class C implments BC, Comparable<BC> { ... }
and changle your list declaration and creation to: ArrayList<BC> list = new ArrayList<>();
EDIT:
To truly guarantee that ArrayList<BC>
is sortable, we need to change the BC
's declaration to:
interface BC extends Comparable<BC> { }
Otherwise we could have class D implements BC
but not Comparable<BC>
. With this new annotation we say "anything that implements BC
must implement Comparable<BC>
as well".
Now you can write only
class B implments BC { ... }
class C implments BC { ... }
(Also note that you can add getString()
method to interface BC
to avoid the instanceof
testing and casting in the compare method)
In order for classes A
and B
to be mutually comparable, these requirements need to be satisfied:
compareTo
on an instance of A
passing an instance of B
must be allowedcompareTo
on an instance of B
passing an instance of A
must be alloweda.compareTo(b)
returns x
, then b.compareTo(a)
must return a value y
with the opposite sign, or zero, when x
is zero.The classes in your code are not mutually comparable, because an attempt to pass an instance of c
to b
's compareTo
(and vice versa) works fine. However, they are not comparable to instances of their own classes, which would create a problem if you were to add more items to the collection to be sorted.
In order for the container to be sortable, your class needs to be comparable to instances of its own type as well.