Reference data types in Java - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a String variable named greeting.

Java

String greeting = [1];

Drag options to blanks, or click blank then click option'

A"Hello"

BHello

Cchar[]

Dint

Attempts:

3 left

💡 Hint

Common Mistakes

Forgetting the quotes around the text.

Using single quotes instead of double quotes.

In Java, String values must be enclosed in double quotes.

2fill in blank

medium

Complete the code to create a new Integer object with value 10.

Java

Integer number = [1];

Drag options to blanks, or click blank then click option'

Anew Integer(10)

BInteger(10)

C10

Dint(10)

Attempts:

3 left

💡 Hint

Common Mistakes

Trying to call the constructor without new.

Assigning a primitive int directly to an Integer without autoboxing.

To create an Integer object, use the new keyword with the constructor.

3fill in blank

hard

Fix the error in the code to compare two String objects correctly.

Java

if (str1.[1](str2)) {
    System.out.println("Strings are equal");
}

Drag options to blanks, or click blank then click option'

A==

Bequals

Cequals()

DcompareTo()

Attempts:

3 left

💡 Hint

Common Mistakes

Using == to compare Strings, which compares references, not content.

Using equals without parentheses, which is a syntax error.

Use the equals() method to compare the contents of two String objects.

4fill in blank

hard

Fill both blanks to create an array of Strings and assign a value to the first element.

Java

String[] names = new String[[1]];
names[[2]] = "Alice";

Drag options to blanks, or click blank then click option'

A3

B0

C1

Dnew String[3]

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 as the first index instead of 0.

Confusing the array size with the index.

The array size is 3, and the first element index is 0.

5fill in blank

hard

Fill all three blanks to create a HashMap, add a key-value pair, and retrieve a value.

Java

Map<String, Integer> map = new [1]<>();
map.[2]("age", 25);
int age = map.[3]("age");

Drag options to blanks, or click blank then click option'

AHashMap

Bput

Cget

DTreeMap

Attempts:

3 left

💡 Hint

Common Mistakes

Using TreeMap instead of HashMap (both work but only one is correct here).

Using get to add entries or put to retrieve values.

Use HashMap to create the map, put to add entries, and get to retrieve values.

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