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.