When is `super()` Really Necessary in a Constructor?

Super() in Constructor: A Question of Necessity

In the realm of object-oriented programming, constructors play a crucial role in initializing newly created objects. However, the question arises: is it essential to explicitly call the super() method within a subclass's constructor?

Automatic Invocation

In the absence of an explicit super() call, the compiler automatically invokes the accessible no-args constructor in the superclass. This constructor has no parameters and is vital for properly initializing the superclass's instance members.

Constructors with Arguments

When a subclass's constructor includes arguments, it typically delegates initialization tasks to a constructor in the superclass with a matching argument list. If no matching constructor is found, an error occurs. In other words, the constructor invoked via super() must have arguments that exactly match the arguments passed to the subclass's constructor.

Importance of Accessibility

The accessible no-args constructor in the superclass must be visible to the subclass. If the superclass constructor is private or protected, an explicit call to super() with appropriate arguments is necessary to invoke the correct constructor.

Examples

To illustrate these concepts, consider the following code snippets:

<code class="java">public class Base { // No-args constructor } public class Derived extends Base { // Default constructor will be invoked by the compiler }</code>

In this case, the default constructor in Base will be automatically invoked when an instance of Derived is created.

<code class="java">public class Base { public Base(int i) { } } public class Derived extends Base { public Derived(int i) { super(i); } }</code>

Here, the constructor in Derived explicitly invokes the matching constructor in Base with the argument i.

<code class="java">public class Base { private Base() { } } public class Derived extends Base { } // Error: no accessible no-args constructor in superclass</code>

This code results in an error because Base's constructor is private and not accessible to the subclass.

In conclusion, while the compiler automatically invokes the accessible no-args constructor in the superclass, explicit super() calls are essential when a subclass's constructor has arguments or when the superclass's constructor is not accessible. Understanding these nuances is crucial for ensuring proper object initialization in Java.

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