Reference data types in Java - Step-by-Step Execution

Concept Flow - Reference data types

Declare reference variable

↓

Assign object to variable

↓

Variable points to object in memory

↓

Use variable to access object data/methods

↓

Change variable to point to another object or null

↓

Old object may be garbage collected if no references

Reference variables hold addresses to objects in memory, not the actual data. They let you access and manipulate objects indirectly.

Execution Sample

Java

String name = "Alice";
String another = name;
name = "Bob";
System.out.println(another);

This code shows how reference variables point to objects and how changing one variable does not affect the other.

Execution Table

Step	Action	Variable 'name'	Variable 'another'	Output
1	Declare and assign "Alice" to name	"Alice" object	null
2	Assign name to another	"Alice" object	"Alice" object
3	Change name to "Bob"	"Bob" object	"Alice" object
4	Print another	"Bob" object	"Alice" object	Alice
5	End	"Bob" object	"Alice" object

💡 Program ends after printing 'Alice', showing 'another' still points to original object

Variable Tracker

Variable	Start	After 1	After 2	After 3	Final
name	null	"Alice"	"Alice"	"Bob"	"Bob"
another	null	null	"Alice"	"Alice"	"Alice"

Key Moments - 2 Insights

Why does printing 'another' show 'Alice' even after 'name' changes to 'Bob'?

Because 'another' points to the original "Alice" object, and changing 'name' to "Bob" makes 'name' point to a new object, not affecting 'another' (see execution_table step 4).

Does assigning one reference variable to another copy the object?

No, it copies the reference (address), so both variables point to the same object initially (see execution_table step 2).

Visual Quiz - 3 Questions

Look at the execution_table at step 3, what does variable 'name' point to?

A null

B "Alice" object

C "Bob" object

D Same as 'another'

💡 Hint

Check the 'Variable name' column at step 3 in execution_table

At which step does 'another' get assigned a value?

A Step 2

B Step 3

C Step 1

D Step 4

💡 Hint

Look at the 'Action' column describing assignment to 'another' in execution_table

If we changed 'name' to point to null at step 3, what would 'another' point to?

A null

B "Alice" object

C "Bob" object

D It would cause an error

💡 Hint

Variable 'another' keeps its reference unless explicitly changed (see variable_tracker)

Concept Snapshot

Reference data types hold addresses to objects, not the data itself.
Assigning one reference to another copies the address, not the object.
Changing one reference variable to a new object does not affect others.
Objects without references can be removed by garbage collection.

Full Transcript

In Java, reference data types store addresses pointing to objects in memory. When you assign one reference variable to another, both point to the same object. Changing one variable to point to a new object does not change the other. For example, if 'name' points to "Alice" and 'another' is assigned 'name', both point to the same "Alice" object. Changing 'name' to "Bob" makes 'name' point to a new object, but 'another' still points to "Alice". This shows how references work differently from simple data copies.

Practice

1.

Which of the following is a reference data type in Java?

int, double, String, boolean

easy

A. String

B. int

C. double

D. boolean

Solution

1. Step 1: Identify primitive vs reference types

   int, double, and boolean are primitive types storing actual values.

2. Step 2: Recognize String as a reference type

   String stores a reference to an object containing text data.

3. Final Answer:

   String -> Option A

4. Quick Check:

   Reference type = String [OK]

Hint: Remember: String is an object, others are primitives [OK]

Common Mistakes:

• Confusing String as primitive
• Thinking int is a reference type
• Assuming all data types are reference types

2.

Which of the following is the correct way to create a new String object in Java?

?

easy

A. String s = new string("hello");

B. String s = new String("hello");

C. String s = String("hello");

D. String s = new String["hello"];

Solution

1. Step 1: Check correct constructor syntax

   Use 'new' keyword with class name and parentheses: new String("hello")

2. Step 2: Verify case sensitivity and brackets

   Class names are case-sensitive (String, not string), and arrays use [] differently.

3. Final Answer:

   String s = new String("hello"); -> Option B

4. Quick Check:

   Correct object creation syntax = String s = new String("hello"); [OK]

Hint: Use 'new ClassName()' with exact case for objects [OK]

Common Mistakes:

• Using lowercase 'string' instead of 'String'
• Missing 'new' keyword
• Using array brackets [] incorrectly

3.

What will be the output of the following code?

String a = "hello";
String b = a;
b = b + " world";
System.out.println(a);

medium

A. world

B. hello world

C. hello

D. Compilation error

Solution

1. Step 1: Understand reference assignment

   Variable b initially points to the same String object as a ("hello").

2. Step 2: String concatenation creates new object

   b = b + " world" creates a new String object; a still points to original "hello".

3. Final Answer:

   hello -> Option C

4. Quick Check:

   Strings are immutable; original stays same [OK]

Hint: String concatenation creates new object, original unchanged [OK]

Common Mistakes:

• Assuming b changes a's value
• Thinking references share mutable data
• Expecting combined string in a

4.

Find the error in the following code snippet:

String[] arr = new String[3];
arr[0] = "Java";
arr[1] = "Python";
arr[3] = "C++";

medium

A. No error, code runs fine

B. Syntax error in array declaration

C. Missing initialization of array elements

D. Array index out of bounds error

Solution

1. Step 1: Check array size and indices

   Array size is 3, valid indices are 0, 1, 2.

2. Step 2: Identify invalid index usage

   arr[3] accesses index 3 which is outside valid range, causing runtime error.

3. Final Answer:

   Array index out of bounds error -> Option D

4. Quick Check:

   Array indices start at 0, max index = size-1 [OK]

Hint: Array indices go from 0 to length-1 only [OK]

Common Mistakes:

• Using index equal to array length
• Confusing syntax error with runtime error
• Assuming arrays auto-expand

5.

Given the class below, what will be the output?

class Box {
  int length;
  Box(int length) { this.length = length; }
}

public class Test {
  public static void main(String[] args) {
    Box b1 = new Box(5);
    Box b2 = b1;
    b2.length = 10;
    System.out.println(b1.length);
  }
}

hard

A. 10

B. 5

C. Compilation error

D. Runtime error

Solution

1. Step 1: Understand reference assignment for objects

   b2 is assigned the reference of b1, so both point to the same Box object.

2. Step 2: Modify object property via one reference

   Changing b2.length to 10 changes the single Box object's length property.

3. Step 3: Print property via original reference

   Printing b1.length shows 10 because b1 and b2 refer to the same object.

4. Final Answer:

   10 -> Option A

5. Quick Check:

   Object references share same data; change reflects everywhere [OK]

Hint: Object references point to same data; changes affect all refs [OK]

Common Mistakes:

• Thinking b1 and b2 are separate objects
• Expecting original value after change
• Confusing primitives with objects