CODEWITHSUNDEEP
javainheritancejava-7
miversen33
miversen33

Reputation: 569

Type Inheritance Issue

I don't really know how to explain the problem I have. Below is an example of a few objects I have.

Lets say I have classes

public class Foo{}
public class FooBar extends Foo{}

All of that is fine, however if I take a list and say I want it to be full of Foo objects, I cannot provide it a FooBar object. Why is this?

Example:

List<Foo> fooList = Arrays.asList(new FooBar());

If FooBar extends Foo, why cant it be cast back to Foo?

Upvotes: 3

Views: 93

Answers (2)

Pshemo
Pshemo

Reputation: 124235

As already mentioned in comments Arrays.asList(T...) returns List<T>. Since you passed FooBar instance to that method type inference from Java 7 used FooBar type as T representation which means that this method returns List<FooBar> which can't be assigned to List<Foo> (you can read more about it at: Is List<Dog> a subclass of List<Animal>? Why aren't Java's generics implicitly polymorphic?)

In Java 8 type inference was improved to make lambdas more useful. Now T infers type not only from parameters, but also from target - reference variable which should hold result. This allows T to infer Foo from List<Foo> fooList rather than new FooBar() argument.

To solve this problem in earlier versions like in your case Java 7 you can explicitly set generic type for generic method asList like

List<Foo> fooList = Arrays.<Foo>asList(new FooBar());
//                         ^^^^^ -- set generic types before method name

More info at: https://docs.oracle.com/javase/tutorial/java/generics/genTypeInference.html (especially "Target Types" section)

Upvotes: 5

AJNeufeld
AJNeufeld

Reputation: 8695

Arrays.asList(new FooBar()) returns a List<FooBar>, which cannot be assigned to a List<Foo>. If it could be assigned to that, then you could create a List<FooBar>, assign it to a List<Foo>, and then add Foo to a list that must contain FooBar, which is clearly an error.

Instead, you need:

List<Foo> fooList = new ArrayList<>();
fooList.add(new FooBar());

Upvotes: 3

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