在java中的超类上调用抽象方法是否有效

Question

..and if so what is the behavior? I came across this in some code I was looking at recently, and it is very confusing to me. I don't have a java compiler, so I can't answer this easily myself. Here is the rough example of what I'm talking about. I would expect this result in a compile error, but as far as I know it is from a working code base.

abstract class Base {
    ...
    abstract boolean foo(String arg);

}

class Sub extends Base {
    ...
    boolean foo(String arg) {
        if(condition) 
            return true;
        else 
            return super.foo(arg); //<-- <boggle/>
    }
}

Answer 1

No, if it's abstract in the superclass you can't call it. Trying to compile your code (having fixed the others) gives this error:

Test.java:13: abstract method foo(String) in Base cannot be accessed directly
            return super.foo(arg); //<-- <boggle/>
                        ^

Answer 2

When you put 'super.' before the method name, you say to compiler: 'hey man! call the method implemented exactly in the Base class'. It doesn't exist there actually so it cannot be called and compiler complains. Just remove 'super.' and leave 'foo(arg);' only. This which will tell the compiler to look for a implementation in some subclass.

BTW, if condition in your example is always false, it'll get into infinitive loop and crash because of out of memory :)

Cheers, ~r

Answer 3

That won't compile. You can't invoke an abstract method.

Answer 4

Abstract method can't be called as it is just a declaration type, without a definition there is no point calling it. Thus Compile time Exception will occur

Answer 5

Tossing your example into Eclipse and editing it so it actually compiles that far produces this error:

"Cannot directly invoke the abstract method foo(String) for the type Base"

Are you sure that comes from a "working code base?"