super protected java

Question

A question about inheritance in java...

class Base {
    private int val = 10;
}

class Derive extends Base{
    public void setVal(int value) {
        super.val = value;
    }
}

Since we can change the private field in super class using super keyword in the subclass, why should we use protected to declare fields in super class?

first of all, as the answer states this won't compile, you can't access private fields, second even if you could, please use setters/getters even internally. — Samy Vilar, Jun 16 '12 at 07:54

score 4 · Accepted Answer · answered Jun 16 '12 at 07:41

4

You can't do that. The code you've given doesn't compile, unless Derive is declared as a nested class within Base (which is a pretty rare case).

You should be getting an error like this:

error: val has private access in Base

answered Jun 16 '12 at 07:41

Jon Skeet

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yes, thanks. I used nested static class for test. It supposed that I should never do this to test accessibility again. – alaska Jun 17 '12 at 18:21

saurabh kumar · Answer 2 · 2014-06-04T03:09:15.620

0

super is an reference variable which is used to call parents constructor.

edited Jun 04 '14 at 03:09

answered Jun 03 '14 at 17:23

saurabh kumar

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5
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super is most definitely not a variable. – matiash Jun 04 '14 at 03:07
@matiash sorry reference variable – saurabh kumar Jun 04 '14 at 03:09

score 0 · Answer 3 · answered Jun 03 '14 at 17:29

0

Check your code you can never access private out side the class even if you have inherited that class.

answered Jun 03 '14 at 17:29

saurabh kumar

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5
26

super protected java

3 Answers3