Generic parameter of a generic class constructor

Asked Oct 17 '15 at 17:37

Active Oct 17 '15 at 17:37

Viewed 48 times

Why can I compile this code without errors:

public class Base<T> {
    Base(List<String> list){}
}

class Child extends Base{
    Child(){
        super(new ArrayList<Integer>());
    }
}

But at the same time I can't compile this one (pay attention to Base<T>):

public class Base {
    Base(List<String> list){}
}

class Child extends Base{
    Child(){
        super(new ArrayList<Integer>());
    }
}

Obviously, there is "incompatible types" in the second example, but why is there no errors in the first one? It is unsafe and can lead to runtime exceptions, can't it?

asked Oct 17 '15 at 17:37

peremeykin

4

In your first example, you're extending a raw type, therefore all use of generics is erased. – Sotirios Delimanolis Oct 17 '15 at 17:41
Excellent question. I see a ```Note: Base.java uses unchecked or unsafe operations. Note: Recompile with -Xlint:unchecked for details.``` when I compile the first example, so it's not exactly without errors. – Siddhartha Oct 17 '15 at 17:42
`ArrayList` doesn't extend `List` – Andrew Tobilko Oct 17 '15 at 17:44
@AndrewTobilko, and what about the first example? – peremeykin Oct 17 '15 at 17:48
@eremeykin, in first example, when you use a raw type, all generic types will be lost. – Andrew Tobilko Oct 17 '15 at 17:59
1

@eremeykin, in second case, `super(new ArrayList());` – Andrew Tobilko Oct 17 '15 at 18:05

Generic parameter of a generic class constructor

0 Answers0