CODEWITHSUNDEEP
javagenerics
Jericho
Jericho

Reputation: 10953

Multiple Bounded Generics in Java

Am having Class A ,Interfaces B,C . aA extends A,aB implements B,cC implements C.But if I try to use MultipleBounded Generics.It is showing compilation error using jdk 1.7. Please help me to correct the errornous. LINK FOR REFERENCE

public class MultipleBounds {
    public static void main(String args[]) 
    {
      D d1 = new D();
      d1.print(new bB()); // compilation Error
      d1.print(new cC()); // compilation Error
      d1.print(new aA()); // It works no Issues
    }
}

class A {
    public void method() {
        System.out.println("methodA");
    }
}

interface B {
    public void method();
}

interface C {
    public void method();
}

class aA extends A {
    public void method() {
        System.out.println("methodaA");
    }
}

class bB implements B {
    public void method() {
        System.out.println("methodbB");
    }
}

class cC implements C {
    public void method() {
        System.out.println("methodcC");
    }
}

class D<T extends A & B & C> 
{
    public void print(T t)
    {
        t.method();
    }
}

Upvotes: 0

Views: 93

Answers (2)

assylias
assylias

Reputation: 328795

The problem is here: D d1 = new D();. You are using a raw type, therefore losing the generics information. And the type erasure of a type intersection A & B & C is its first component (A). So your d1 can only accepts As.

For example, if you reorder it to B & A & C, you will only be able to pass Bs.

Finally note that you would not be able to declare your D using generics because no class in your code is an A and a B and a C. You could create a class ABC extends A implements B, C and use D<ABC> d = new D<>();

Upvotes: 2

Rohit Jain
Rohit Jain

Reputation: 213351

Since you are using raw type D for invoking the method, all the generic information from the class is lost. And the method print will be replaced by it's erasure.

Now, the erasure of the type parameter T is the left-most bound it has, which is A. This is evident in JLS §4.6:

The erasure of a type variable (§4.4) is the erasure of its leftmost bound.

So, what compiler sees is this:

class D {
    public void print(A t)
    {
        t.method();
    }  
}

That is why, it works for A and not for B and C. Lesson: Avoid using raw types.

Upvotes: 3

Related Questions