Passing Arguments in Java – Value & Reference

Java handles data passing in a unique but consistent way. This page explains one of the most misunderstood topics: Pass by Value and Objects as Parameters.

1. Passing Arguments by Value

Java is strictly pass-by-value. This means a copy of the variable is passed to methods. Modifying the copy does NOT affect the original value.

class Test {
    void change(int x) {
        x = 50;
    }

    public static void main(String[] args) {
        int a = 10;
        new Test().change(a);
        System.out.println(a);
    }
}

Output: 10 (Original value unchanged)

2. Class Objects as Parameters (Looks Like Reference)

Objects behave differently because the value passed is a copy of the reference. Both references point to the same object, so modifying the object reflects outside the method.

class Car {
    int speed;
}

class Demo {
    void modify(Car c) {
        c.speed = 100;
    }

    public static void main(String[] args) {
        Car car = new Car();
        car.speed = 40;

        new Demo().modify(car);
        System.out.println(car.speed);
    }
}

Output: 100 (Object modified!)

Java does NOT pass objects by reference. It passes a copy of the reference → but the referenced object is the same.

Practice Questions – Click to Reveal Answers

1️⃣ What does Java pass: value or reference?

Java always passes by value. For objects, it passes a copy of the reference.

2️⃣ Why does modifying an object inside a method reflect outside?

Because the reference copy still points to the same object in memory.

3️⃣ Will reassigning an object inside a method affect the original?

No. Reassignment changes only the local copy of the reference.

Quick Summary

Java passes everything by value (copies).
Primitive copies do NOT affect originals.
Object reference copies point to the same object.
Modifying object fields affects the original object.
Reassigning references affects only the local copy.