Why is cast necessary for calling methods inherited from interfaces?

Question

I was practicing to program for interface in java when I came across something I could not understand. Let me call the class Country

 class Country implements Comparable
    {
    int a; 
    int b; 
    int c;
    public Country(int _a,int _b,int _c)
    {
    a=_a;
    b=_b;
    c=_c;
    }
    public int compareTo(Object obj)
    {
    /*compares a, b and c and returns number of variables greater for this object.I am not include instanceof check*/
    int count=0;
    Country other=(Country)other;
    if(a>other.a)
    count++;
    else 
    if(a<other.a)
    count--;
    if(b>other.b)
    count++;
    else 
    if(b<other.b)
    count--;
    if(c>other.c)
    count++;
    else 
    if(c<other.c)
    count--;
    return count;
    }
public void write()
{
System.out.println(" hello");
}
    public static void main(String args[])
    {
    Object p=new Country(1,2,3);
    Object q=new Country(2,3,4);
    System.out.println(p.compareTo(q));
    }
    }

So the question here is if we declare something as

Object p=new Country(1,2,3);
Object q=new Country(2,3,4); 
p.write();

This works. but why not

p.compareTo(q)//as done in the main code

Why is this cast required?

((Comparable)p).compareTo(q);

Note: You should be using `class Country implements Comparable`—avoid [raw types](https://stackoverflow.com/questions/2770321/what-is-a-raw-type-and-why-shouldnt-we-use-it). — Slaw, Jun 24 '19 at 08:11
`p.write(); ` does **not** work neither, at least for me. Not sure what you've done for it to work. — Amongalen, Jun 24 '19 at 08:14
Also, I strongly doubt that your implementation of `compareTo` meets the [formal requirements](https://docs.oracle.com/javase/8/docs/api/java/lang/Comparable.html) of the `Comparable` interface — Andy Turner, Jun 24 '19 at 08:14
@Slaw I do know about the generic interface but my question is about this one in particular. — dawood mansoor, Jun 24 '19 at 08:19
@Andy Turner yes it does not because it does not follow x.compareTo(y)=-y.compareTo(x); — dawood mansoor, Jun 24 '19 at 08:20
@Andy Turner yes i do not feel it is a duplicate . please request the removal of duplicacy. — dawood mansoor, Jun 24 '19 at 08:21
@GhostCat yes the question of which has been a duplicate answers my question. Oh it is just so difficult to ask a non duplicate question. — dawood mansoor, Jun 24 '19 at 08:32
@dawoodmansoor "Oh it is just so difficult to ask a non duplicate question" that means that you're not alone in things you find confusing :) It's sometimes hard to know what to search for, though, and those of us who have been around here for a long time can find it easily. — Andy Turner, Jun 24 '19 at 08:36
@AndyTurner I realized that when I saw the headings of the above questions. I will get used to it , god willing. — dawood mansoor, Jun 24 '19 at 08:40

score 2 · Answer 1 · answered Jun 24 '19 at 08:08

2

Because you are storing the value in a reference of type Object, so, the compiler has no way to know that it is actually a Country and that it implements Comparable at all.

answered Jun 24 '19 at 08:08

lealceldeiro

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score 2 · Accepted Answer · answered Jun 24 '19 at 08:11

Here:

Object p=new Country(1,2,3);

The compiler could know that p is of type Country. But it doesn't know it.

It only sees that you declared p to be of type Object. And the class Object does not have a print() method, or a compareTo() method.

Polymorphism (looking up methods on the actual type of objects) happens at runtime, but deciding whether some object has a specific method happens at compile time. And in this case, the compiler decides: p is an Object, thus it is missing the method you want to invoke!

Why is cast necessary for calling methods inherited from interfaces?

2 Answers2