内部类构造函数调用

Question

The JLS says:

The constructor of a non-private inner member class implicitly declares, as the first formal parameter , a variable representing the immediately enclosing instance of the class.

---

Ok, if we write the following:

class A:

package org.gradle;

public class A extends B.Inner{

    public A(B b){
        b.super();        //OK, invoke B.inner(B)        
    }
}

class B:

package org.gradle;

public class B{

    public class Inner{
    }
}

As said here , b.super() actually invoke B.Inner(B) .

---

But if we write

class B:

package org.gradle;

public class B {
    class Inner{ 
        public Inner(B b){
            System.out.println("Inner(B)");
        }
    }
}

class A:

package org.gradle;

public class A extends B.Inner{

    public A(B b) {
        b.super(); //The constructor B.Inner() is undefined
    }
}

So, in the latter example b.super() tries to invoke B.Inner() instead. Why is that so difference?

Answer 1

It does try to invoke B.Inner(B) in your second example.

It can't find it because there's only a B.Inner(B, B) . If your constructor is B.Inner() then it gets changed to Inner(B) ... and if your constructor is B.Inner(B) then it gets changed to B.Inner(B, B) .

Note that the hidden parameter is effectively an implementation detail, and unless you're studying how the Java compiler works, you don't need to know it exists.