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The following code does not give any error : 

List<A> l = new LinkedList<A>();
l.add(new B()); // B extends A

Then why does the below give error:

List<B> l = new LinkedList<B>();
List<A> l1 = l;
ghostrider
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2 Answers2

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B extends A does not mean List<B> extends List<A>. Assigning sub type collection to parent type will make it not type safe.

List<B> ListB = new LinkedList<B>();
List<A> ListA = ListB ;  // Suppose you can compile it without error
ListA.add(new C());  // You can put C into ListA if C extends A
B b = ListB.get(0); // Then when you retrieve from ListB, you get a C, not type safe!

You need a wildcard to guarantee type safe and make it compile:

List<B> ListB = new LinkedList<B>();
List<? extends A> ListSubA = ListB ;
ListSubA.add(new C()); // Compile error, a wildcard can prevent user putting C into it
ListB.add(new B());  // You can only add new element by ListB 
B b = ListB.get(0);  // Type safe
marc_s
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xingbin
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1

It is explained here - https://docs.oracle.com/javase/tutorial/java/generics/inheritance.html

Basically List<A> is treated as a separate (not related in any way) type from List<B> so when you declare a reference of type List<A> it cannot point to an object of type List<B>.

There is a nice discussion here - Is List<Dog> a subclass of List<Animal>? Why are Java generics not implicitly polymorphic?

neta
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