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Why erasure of generic types prevents overriding in this case?

The code below doesn't compile on OpenjDK 11. It seems to me that test1 in B should override test1 in A, because:

  1. The methods have same name.
  2. The methods have same parameter list.
  3. The methods have same visibility.
  4. The methods actually do not throw potentially incompatible checked exceptions.
  5. Their return types are covariant.

I took a decompiler and decompiled each class separately. The compiled code actually worked as I would expect. It replaced U extends Number with Number, T extends Integer with Integer and so on. But when I try to compile the two classes together, I got and error on the second class that says the test in the second class does not override a method in the first one.

I'm missing something big or small here. And it's probably related to 5. Maybe the types are not covariant. Can you help me?

class A {
    <U extends Number, T extends Number> U test(T test) {
        System.out.println("In A.test(T test)");
        return null;
    }
    //Decompiler shows that the above method erases to 
    // Number test(Number test)
    // Just like the method below.
        
    Number test2(Number test) {
        return null;
    }
}

class B extends A {
    //Unsuccessful override. Compiler error.
    @Override  
    <U extends Integer, T extends Number> U test(T test) {
        System.out.println("In B.test(T tesT)");
        return null;
    }
    //Decompiler shows that the above method erases to 
    //Integer test(Number test)
    //Just like the method bellow.
    
    
    //Successful override
    @Override
    Integer test2(Number test) {
        return null;
    }
}
over 4 years ago · Santiago Trujillo
1 answers
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0

The reason why your test methods produce an error is because they have totally unrelated, different signatures. Note that the signature of a method consists of its name, parameter list, and type parameters.

Quote from the Java Language Specification:

Two methods or constructors, M and N, have the same signature if they have the same name, the same type parameters (if any) (§8.4.4), and, after adapting the formal parameter types of N to the type parameters of M, the same formal parameter types.

Crucially, your two test methods do not have the same type parameters, because U in A.test has a different bound from U in B.test.

Two methods or constructors M and N have the same type parameters if both of the following are true:

  • M and N have same number of type parameters (possibly zero).

  • Where A1, ..., An are the type parameters of M and B1, ..., Bn are the type parameters of N, let θ=[B1:=A1, ..., Bn:=An]. Then, for all i (1 ≤ i ≤ n), the bound of Ai is the same type as θ applied to the bound of Bi.

Think about what would happen if B.test actually overrode A.test. You could pass a type to the type parameter U that is out of its bounds!

A a = new B();
// This will call B.test, and U is Double, T is Integer
// but U should extends Integer!
Double x = a.test((Integer)0); 

For more info, here are the precise rules for when overriding happens. Note that criteria #4 and #5 on your list are not actually considered. They are just additional requirements that make your code not compile if you break them. One method is still "defined" to override another even if you break those requirements. They are listed here in the JLS.

over 4 years ago · Santiago Trujillo Report
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