CODEWITHSUNDEEP
javagenerics
whiteErru
whiteErru

Reputation: 275

Java (1.6) generics wildcard

Consider flowing situation:

    class A {

    }

    class B extends A {

    }


    List <? extends A> x = new ArrayList<A>();
    List <? extends A> xx = new ArrayList<B>();

both 'x' and 'xx' are legal declarations in Java 6 and Java 7 (I know that in Java 7 you can also substitute the parameterized type of the constructor with an empty set of type parameters (<>). However, I wonder, what is the difference between 'x' and 'xx' in Java 6?

Upvotes: 5

Views: 316

Answers (2)

Peter Perh&#225;č
Peter Perh&#225;č

Reputation: 20782

no difference. it all boils down to two lists of objects. You are just trying to coax the compiler to lend you a helping hand, as it's quite clever and could help you catch bad usage. You tell it that as long as the type in the List is-a A, all is fine. The compiler can then help highlighting any problems.

Upvotes: 1

SLaks
SLaks

Reputation: 887415

The way you wrote it, there is no difference.

At compile-time, both have the same type; namely, a List<> of some unknown type that extends A.
This is why you can't add anything to either list – you have no idea whether it's actually a List<B> or a List<C> or of some other type.

At runtime, they also have the same type; namely, List. Due to type erasure, the type parameter doesn't exist at runtime.

Because you don't save any more strongly-typed references to it, no-one can tell that one of the lists is actually a List<B>.

The point of these wildcards is for functions.
You can make a function that takes a List<? extends A>, then pass it a List<B> or a List<C>, and the calling code can keep using the original list with its original type.

Upvotes: 6

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