I hope this code explains the problem:
class Foo {
void a() { / *stuff */ }
}
class Bar extends Foo {
void a() { throw new Exception("This is not allowed for Bar"); }
class Baz {
开发者_开发技巧 void blah() {
// how to access Foo.a from here?
}
}
}
I know that I may be doing something wrong, because inheritance perhaps shouldn't be used in such way. But it's the easiest way in my situation. And, beside that, I'm just curious. Is it possible?
Bar.super.a()
appears to work.
Per JLS section 15.12
ClassName . super . NonWildTypeArguments_opt Identifier ( ArgumentList_opt )
is a valid MethodInvocation
You can call any method from the outer class with Outer.this.method()
.
But methods are resolved at runtime, so if you have overridden it in your subclass, only the subclass method (Bar.a()
) can access the original (by calling super.a()
).
As you probably discovered, you can't write Bar.this.super.a()
-- but even if you could, it would still give you Bar.a()
, not Foo.a()
.
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