Introduction

In Java, this() and super() are not ordinary method calls. They are special constructor-invocation statements, and they must appear first because Java needs one clear, valid initialization path before ordinary constructor code starts touching the object.

What this() and super() Actually Do

super() calls a constructor in the direct superclass. this() delegates to another constructor in the same class.

1class Animal {
2    Animal(String type) {
3        System.out.println(type);
4    }
5}
6
7class Dog extends Animal {
8    Dog() {
9        super("dog");
10    }
11}

Constructor chaining with this() looks like this:

1class Point {
2    private final int x;
3    private final int y;
4
5    Point() {
6        this(0, 0);
7    }
8
9    Point(int x, int y) {
10        this.x = x;
11        this.y = y;
12    }
13}

In both cases, Java is choosing the construction path before any normal constructor-body logic runs.

Why the Call Must Come First

Java initializes an object from the top of the inheritance chain downward. Before subclass code can safely rely on inherited state, the superclass part of the object must be constructed.

Likewise, when one constructor delegates to another through this(), Java must jump to that target constructor immediately so there is one consistent path through the class's initialization logic.

If arbitrary statements could come first, several problems would appear:

• subclass code could run before the superclass state was initialized,
• two constructors in the same class could both start doing work before the actual constructor chain was decided,
• final-field initialization rules would be harder to enforce,
• the compiler would lose a simple, predictable model of object construction.

The rule is restrictive on purpose. It protects the object from being used in a partially initialized state.

Java Inserts super() Automatically

If you do not write this() or super() explicitly, the compiler inserts a no-argument super() call when possible.

1class Base {
2    Base() {
3        System.out.println("base");
4    }
5}
6
7class Child extends Base {
8    Child() {
9        System.out.println("child");
10    }
11}

This behaves as though the constructor started with super();.

That is also why compilation fails if the superclass has no no-argument constructor and you forgot to call one of its available constructors explicitly.

You Cannot Call Both

A constructor can use either this() or super() as its first statement, but not both. If you call this(), the target constructor eventually becomes responsible for reaching super().

That keeps the construction path linear and unambiguous. There must be one chain, not several competing initialization branches.

Why This Helps the Compiler and the Reader

The rule is not only about runtime safety. It also makes constructor behavior easier to reason about.

When you read a constructor, you immediately know:

• whether it delegates locally with this(),
• whether it invokes a superclass constructor directly,
• that any normal statements run only after the constructor chain has been chosen.

That predictability helps both the compiler and the human reader.

Common Pitfalls

A common mistake is thinking of this() and super() as ordinary methods. They are special constructor-invocation syntax with stronger placement rules.

Another issue is forgetting that Java inserts super() automatically when neither constructor call is written. That hidden insertion explains many compilation errors involving superclass constructors.

Developers also sometimes spread too much logic across many chained constructors, which stays legal but becomes hard to follow.

Summary

• 'this() delegates to another constructor in the same class.'
• 'super() invokes a constructor in the superclass.'
• Both must come first so Java can establish one valid initialization path.
• The rule protects objects from partial initialization and keeps constructor order predictable.
• If you omit both, Java inserts an implicit no-argument super() when it can.