When extending a gridview, why need to call super in constructor?

Question

I implemented an expandedGridView like this Kishanjvaghela's github example or the answer of Raj008 in stack Overflow.

It works fine, but I wonder why does it need to call super() in the constructor

public class ExpandableHeightGridView extends GridView {
    boolean expanded = false;

    public ExpandableHeightGridView(Context context) {
        super(context);
    }

    public ExpandableHeightGridView(Context context, AttributeSet attrs) {
        super(context, attrs);
    }

    public ExpandableHeightGridView(Context context, AttributeSet attrs, int defStyle) {
        super(context, attrs, defStyle);
    }
/* more code here */
}

If there is no constructor, java must call parent constructor, isn't it ?

Check this [Why is constructor of super class invoked when we declare the object of sub class?](https://stackoverflow.com/q/7173019/7666442) and this one [Why do this() and super() have to be the first statement in a constructor?](https://stackoverflow.com/a/6318640/7666442) — AskNilesh, Jul 08 '19 at 08:15
extending a class, you have to declare at least one constructor calling parent class constructor. Only if a parent class has the default constructor (with no arguments) you can skip declaring constructor. — Vladyslav Matviienko, Jul 08 '19 at 08:25

score 0 · Accepted Answer · answered Jul 08 '19 at 08:22

0

Because by default(optional) super() keyword will call no-arg constructor, So to call argumented constructor of parent class it must to write super() keyword with arguments

just see how constructor chaining works

answered Jul 08 '19 at 08:22

Rahul

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When extending a gridview, why need to call super in constructor?

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