Type mismatch with generics
Here's an interface:
public interface Foo<T> extends Comparable<Foo<T>> {
...
}
And there are some classes implementing this interface:
public class Bar extends Something implements Foo<Something> {
public Vector<Foo<Bar>> giveBar() {
...
}
}
public class Boo extends SomethingElse implements Foo<SomethingElse> {
public Vector<Foo<Boo>> giveBoo() {
...
}
}
Now I want to keep a bunch of Foos (that may really be Foos or Boos) inside a vector.
Bar bar = new Bar();
Boo boo = new Boo();
Vector<Foo<?>> vector;
if (...)
vector = bar.giveBar();
else
vector = boo.giveBoo();
I get:
Type mismatch: cannot convert from Vector<Foo<SomethingElse>> to Vector<Foo<?>>
The same goes for:
Vector<Foo> vector;
if (...)
vector = giveBar();
else
vector = giveBoo();
Is a superclass that both Bar and Boo extend the only solution to this problem?
Asked by: Grace777 | Posted: 21-01-2022
Answer 1
What all that code boils down to is:
Vector<A> vector = new Vector<B>();
In this case B extends A, but that's not allowed because the types don't match. To make clear why this doesn't work, imagine the following code:
Vector<Vector<?>> vector = new Vector<Vector<String>>();
vector.add(new Vector<Integer>());
The variable's type is of a vector of vectors of unknown type; and what's being assigned to it is a vector of vectors of strings. The second line adds a vector of integers to that. The component type of the variable Vector<?>, which accepts Vector<Integer>; but the actual vector's component type is Vector<String>, which doesn't. If the compiler didn't object to the assignment on the first line, it would allow you to write the incorrect second line without being spotted.
C#'s generics have a similar restriction, but the difference is that a generic class in C# stores it component type, while Java forgets component types when the code is compiled.
ps - Why on earth are you using Vector rather than LinkedList or ArrayList? Is it because there are threading issues involved?
Answer 2
You can use
Vector<? extends Foo<?>> vector;
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