CODEWITHSUNDEEP
javaoopreferencepolymorphismcompile-time
rb612
rb612

Reputation: 5563

Are reference objects with different types any different in Java?

When saying:

  String str = "hello";
  Object obj = str;
  System.out.println(str==obj);

The result is true, because it points to the same objects in memory, which makes sense. But if I say:

obj.indexOf("h");

Or any subclass method, I get "cannot find symbol". They're still pointing to the same object, so what's going on during compile-time that makes reference objects of different types different from each other?

Upvotes: 1

Views: 41

Answers (2)

Jan
Jan

Reputation: 13858

Apples and Pears.

The identity check == is performed at runtime. And valid and fine.

The construct obj.indexOf... is a compile time error as class Object just does not have a method indexOf

If you tell the compiler (by means of casting) that obj contains a String, you can get valid code

((String)obj).indexOf("h");

Will compile again

Upvotes: 0

duffymo
duffymo

Reputation: 308753

The Object type reference only knows about methods that are part of its public interface.

You have to cast if you know that Object reference is a String type:

int index = ((String) obj).indexOf("h");

Upvotes: 3

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