Why Is a Method Executed Before the Default Constructor in Java?

Why does a method run before the default constructor in Java?

class Parent {
    Parent() {
        System.out.println("Parent Constructor");
        method(); // Calling method in constructor
    }
    void method() {
        System.out.println("Method called");
    }
}

class Child extends Parent {
    Child() {
        System.out.println("Child Constructor");
    }
}

public class Test {
    public static void main(String[] args) {
        Child child = new Child();
    }
}

In Java, the execution order in class instantiation can sometimes lead to confusion, especially regarding when constructors and methods are called during object creation. Understanding the initialization phases of Java objects clarifies why a method may execute prior to a default constructor.

class Base {
    Base() {
        display(); // Method call in constructor
    }
    void display() {
        System.out.println("Base Class Method");
    }
}

class Derived extends Base {
    Derived() {
        System.out.println("Derived Class Constructor");
    }
    void display() {
        System.out.println("Derived Class Method");
    }
}

Causes

• Java initializes the parent class before the child class when creating an instance of a child class.
• If a method is called explicitly from within a constructor, it will execute at the point of the method invocation, regardless of the constructor sequence.
• Java does not invoke the parent's default constructor by itself unless it's specified explicitly (via super() or implicitly if no constructor is defined in the child class).

Solutions

• Always ensure proper constructor calls using super() to control the flow of execution clearly.
• Be mindful of method calls in constructors that may lead to unexpected behavior or state before the object is fully constructed.

Mistake: Calling methods within the constructor of a parent class without understanding the subclass's method overrides.

Solution: Be cautious of calling methods within the constructor; if the subclass overrides the method, the overridden version executes, potentially before the subclass constructor runs.

Mistake: Assuming the parent class's constructor completes before any method calls.

Solution: Remember that if a method is invoked in the constructor, the method body executes at that time, which can lead to partially constructed objects being accessed.