Reference data types in Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What are reference data types in Java?

Reference data types store the address (reference) of objects in memory, not the actual data value itself. Examples include classes, arrays, and interfaces.

beginner

How do reference types differ from primitive types in Java?

Primitive types store actual values (like int, double), while reference types store memory addresses pointing to objects. Reference types can be null; primitives cannot.

intermediate

What happens when you assign one reference variable to another?

Both variables point to the same object in memory. Changes through one reference affect the same object seen by the other.

beginner

Can reference variables be null? What does that mean?

Yes, reference variables can be null, meaning they do not point to any object in memory. Accessing methods or fields on a null reference causes a NullPointerException.

beginner

Give an example of a reference data type in Java.

Example:

String name = "Alice";

Here, name is a reference variable pointing to a String object containing "Alice".

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

A int

B double

C String

D boolean

String is a class in Java, so it is a reference data type. int, double, and boolean are primitive types.

What does a reference variable store?

A The actual data value

B The size of the object

C The data type of the object

D The memory address of an object

Reference variables store the memory address (reference) of the object they point to.

What happens if you try to use a method on a null reference variable?

A The program crashes with NullPointerException

B The method returns null

C The variable automatically gets initialized

D The method runs normally

Using a method on a null reference causes a NullPointerException at runtime.

If you assign one reference variable to another, what happens?

A Both variables point to the same object

B A new object is created

C The original object is deleted

D The variables become primitive types

Both reference variables will point to the same object in memory.

Which of these is NOT a reference data type?

A Interface

B char

C Array

D Class

char is a primitive data type, not a reference type.

Explain what reference data types are in Java and how they differ from primitive types.

Think about how variables hold data versus how they hold addresses.

Describe what happens when you assign one reference variable to another in Java.

Consider what happens to the memory address stored.

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