Java Super Keyword

this means "the current object". super means "the parent's part of the current object". A subclass uses it to build its parent part, to reuse the parent's version of a method it has overridden, or to reach a parent field it has hidden with one of the same name. Three uses, one idea: go up one level.

The Three Uses

All Three in One Program

public class SuperDemo {
    static class Account {
        String type = "Account";
        String holder;
        long balance;
 
        Account(String holder, long balance) {
            this.holder = holder;
            this.balance = balance;
        }
 
        String describe() { return holder + " has Rs. " + balance; }
    }
 
    static class SavingsAccount extends Account {
        String type = "Savings";
 
        SavingsAccount(String holder, long balance) {
            super(holder, balance);
        }
 
        @Override
        String describe() {
            return super.describe() + " in a " + type
                    + " account (parent type: " + super.type + ")";
        }
    }
 
    public static void main(String[] args) {
        System.out.println(new SavingsAccount("Srihitha", 12000).describe());
    }
}

Srihitha has Rs. 12000 in a Savings account (parent type: Account)

How it works

1. Line 19: super(holder, balance) runs Account's constructor, which sets up the inherited fields.
2. Line 24: super.describe() calls the parent's describe(), even though the child has overridden it. The child adds to the result instead of repeating the parent's code.
3. The child declares its own type field, which hides the parent's. Plain type means the child's ("Savings"); super.type reaches the parent's ("Account").

Try it yourself: change super.describe() to describe() in line 24. What happens when you run it, and why?

super(...) Rules

The automatic super() causes a common error when the parent has no no-arg constructor (see the next lesson).

Common Mistakes

Calling describe() instead of super.describe() inside the override. The method calls itself forever and ends in StackOverflowError.

Expecting super.super.method(). There's no such thing; super reaches only the direct parent.

Using super in a static method. Like this, it needs an object.

Interview Corner

Practice

1. Easy: Write Manager extends Employee where Manager's constructor calls super(name) and adds a teamSize.
2. Harder: Override toString() in Manager so it returns the parent's toString() result plus the team size, using super.toString().

Key Takeaways