Reference data types in Java - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Java code with String references?

Consider the following Java code snippet. What will be printed when it runs?

Java

public class Main {
    public static void main(String[] args) {
        String a = "hello";
        String b = a;
        a = "world";
        System.out.println(b);
    }
}

A hello

B world

C null

D Compilation error

💡 Hint

Remember that Strings are immutable and variables hold references.

Variable 'b' holds the reference to the original string "hello". Changing 'a' to "world" does not affect 'b'. So printing 'b' outputs "hello".

❓ Predict Output

intermediate

What is the output of this Java code with arrays and references?

Look at this Java code. What will it print?

Java

public class Main {
    public static void main(String[] args) {
        int[] arr1 = {1, 2, 3};
        int[] arr2 = arr1;
        arr2[0] = 10;
        System.out.println(arr1[0]);
    }
}

A Compilation error

B 10

C 0

D 1

💡 Hint

Arrays are objects and variables hold references to them.

Both 'arr1' and 'arr2' refer to the same array object. Changing arr2[0] changes arr1[0]. So output is 10.

🔧 Debug

advanced

What error does this Java code produce?

Examine this Java code. What error will it cause when compiled or run?

Java

public class Main {
    public static void main(String[] args) {
        String s = null;
        if (s.equals("test")) {
            System.out.println("Match");
        }
    }
}

A ArrayIndexOutOfBoundsException

B Compilation error: cannot call method on null

C No output, program runs fine

D NullPointerException at runtime

💡 Hint

Calling a method on a null reference causes an error.

Calling s.equals(...) when s is null causes a NullPointerException at runtime.

🧠 Conceptual

advanced

How many objects are created in this Java code?

Consider this Java code snippet. How many distinct objects are created in memory?

Java

public class Main {
    public static void main(String[] args) {
        String s1 = new String("abc");
        String s2 = "abc";
        String s3 = s1;
    }
}

A 3

B 1

C 2

D 0

💡 Hint

Remember string literals and new String() create objects differently.

The literal "abc" creates one object in the string pool. new String("abc") creates a new distinct object. s3 just references s1, no new object.

❓ Predict Output

expert

What is the output of this Java code with object references and mutation?

Analyze this Java code. What will it print?

Java

class Box {
    int value;
    Box(int v) { value = v; }
}

public class Main {
    public static void main(String[] args) {
        Box b1 = new Box(5);
        Box b2 = b1;
        b2.value = 10;
        b1 = new Box(20);
        System.out.println(b2.value);
    }
}

A 10

B 5

C 20

D Compilation error

💡 Hint

Changing b1 to a new object does not affect b2's reference.

b2 and b1 initially point to the same Box object. Changing b2.value to 10 affects that object. Then b1 points to a new Box(20), but b2 still points to the old object with value 10.

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