Why not super.super.method() in Java?

When I was looking at a question, I thought that using the super.super.method() like the code below would solve it easily.

@Override
public String toString() {
    return super.super.toString();
}

I don't know if super.super.method() is useful in many cases, but I suddenly wondered. Why isn't this working? And can't you do this in other languages? If anyone knows, please explain.

java superclass

2022-09-22 22:28

1 Answers

super.super.method() violates encapsulation. The super.super.method bypasses the processing of the parent class, which creates many problems.

For example, suppose you have a code that collects items as follows:

public class Items
{
    public void add(Item item) { ... }
}

public class RedItems extends Items
{
    @Override
    public void add(Item item)
    {
        if (!item.isRed())
        {
            throw new NotRedItemException();
        }
        super.add(item);
    }
}

public class BigRedItems extends RedItems
{
    @Override
    public void add(Item item)
    {
        if (!item.isBig())
        {
            throw new NotBigItemException();
        }
        super.add(item);
    }
}

It's okay up to here. If you look at BigRedItems, RedItems always have a red color because they are processed by RedItems in super.add(item) and if the item is not red, they will make an exception. And if super.super.add(); is now possible

public class NaughtyItems extends RedItems
{
    @Override
    public void add(Item item)
         {
                  // No exceptions are made whether the item is red or not.
        super.super.add(item);
          }
}

RedItems are not treated as exceptions, so there is no point in inheriting them.

2022-09-22 22:28

If you have any answers or tips

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