Unboxing in Java - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this unboxing example?

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

Java

Integer a = 1000;
int b = a;
System.out.println(b == a);

A true

B Runtime exception

C Compilation error

D false

💡 Hint

Remember that unboxing converts Integer to int automatically.

The Integer 'a' is unboxed to int 'b'. The comparison 'b == a' compares int and Integer, but 'a' is unboxed too, so it compares two ints with the same value, resulting in true.

❓ Predict Output

intermediate

What is the output when comparing Integer objects with ==?

What will this Java code print?

Java

Integer x = 127;
Integer y = 127;
System.out.println(x == y);
Integer m = 128;
Integer n = 128;
System.out.println(m == n);

A

false
false

B

true
true

C

true
false

D

false
true

💡 Hint

Integer caching affects values between -128 and 127.

Java caches Integer objects for values from -128 to 127. So x and y point to the same object, making x == y true. For 128, m and n are different objects, so m == n is false.

🔧 Debug

advanced

Identify the runtime error caused by unboxing

What error will this code produce when run?

Java

Integer a = null;
int b = a;
System.out.println(b);

A NullPointerException

B Compilation error

C Prints 0

D No output, infinite loop

💡 Hint

Unboxing null causes a problem at runtime.

Unboxing tries to convert null Integer to int, which causes a NullPointerException at runtime.

❓ Predict Output

advanced

What is the output of this mixed boxing and unboxing code?

Analyze the output of this Java code snippet:

Java

Integer i = 10;
int j = 10;
System.out.println(i.equals(j));
System.out.println(i == j);

A

false
false

B

true
true

C

true
false

D

false
true

💡 Hint

Remember how equals and == behave with Integer and int.

i.equals(j) returns true because equals compares values after unboxing j. i == j compares Integer and int, unboxing i, so true.

🧠 Conceptual

expert

How many Integer objects are created in this code?

Consider this Java code snippet. How many distinct Integer objects are created during execution?

Java

Integer a = 100;
Integer b = 100;
Integer c = new Integer(100);
Integer d = Integer.valueOf(100);
Integer e = 200;
Integer f = 200;

A 4

B 5

C 3

D 6

💡 Hint

Remember Integer caching and when new objects are created.

a and b use cached Integer for 100 (one object), c creates a new object, d uses cached object again, e and f for 200 create two new objects (no cache). Total 4 objects: one cached for 100, one new from c, two new for 200.

Practice

1. What does unboxing mean in Java?

easy

A. Converting a primitive type to a String

B. Automatically converting a wrapper object to its primitive type

C. Manually converting a primitive type to a wrapper object

D. Automatically converting a String to a wrapper object

Solution

1. Step 1: Understand wrapper and primitive types

   Wrapper classes like Integer wrap primitive types like int.

2. Step 2: Define unboxing

   Unboxing is the automatic conversion from wrapper object to primitive type.

3. Final Answer:

   Automatically converting a wrapper object to its primitive type -> Option B

4. Quick Check:

   Unboxing = wrapper to primitive [OK]

Hint: Unboxing means wrapper to primitive automatically [OK]

Common Mistakes:

• Confusing unboxing with boxing (primitive to wrapper)
• Thinking unboxing converts to String
• Assuming unboxing requires manual code

2. Which of the following Java statements correctly uses unboxing?

easy

A. Integer num = 10; int val = num;

B. Integer val = new Integer(10);

C. Integer num = int 10;

D. Integer val = Integer.valueOf(10);

Solution

1. Step 1: Check each option for correct syntax and unboxing

   Integer num = 10; int val = num; assigns Integer object to int variable, triggering unboxing automatically.

2. Step 2: Identify errors in other options

   Integer val = new Integer(10); assigns wrapper object to wrapper variable (no unboxing); Integer num = int 10; has invalid syntax; Integer val = Integer.valueOf(10); assigns wrapper object to wrapper variable (no unboxing).

3. Final Answer:

   Integer num = 10; int val = num; -> Option A

4. Quick Check:

   Wrapper to primitive assignment triggers unboxing [OK]

Hint: Assign wrapper object to primitive variable to unbox [OK]

Common Mistakes:

• Assigning wrapper constructors directly to wrapper variables (no unboxing)
• Incorrect syntax like 'Integer num = int 10;'
• Confusing boxing and unboxing syntax

3. What is the output of this Java code?

Integer a = 5;
int b = a + 10;
System.out.println(b);

medium

A. NullPointerException

B. 510

C. Compilation error

D. 15

Solution

1. Step 1: Understand unboxing in expression

   Variable 'a' is Integer object; adding 10 triggers unboxing 'a' to int 5.

2. Step 2: Calculate the sum and print

   5 + 10 = 15, so output is 15 printed.

3. Final Answer:

   15 -> Option D

4. Quick Check:

   Unboxing Integer 5 + 10 = 15 [OK]

Hint: Unboxing happens automatically in arithmetic expressions [OK]

Common Mistakes:

• Thinking it concatenates as string '510'
• Expecting compilation error due to wrapper usage
• Confusing with NullPointerException if variable was null

4. Identify the error in this code snippet:

Integer num = null;
int val = num;
System.out.println(val);

medium

A. Prints 0 without error

B. Compilation error due to null assignment

C. Runtime NullPointerException during unboxing

D. Prints 'null' as output

Solution

1. Step 1: Analyze unboxing of null Integer

   Variable 'num' is null; unboxing null to int causes runtime error.

2. Step 2: Identify exception thrown

   Unboxing null throws NullPointerException at runtime.

3. Final Answer:

   Runtime NullPointerException during unboxing -> Option C

4. Quick Check:

   Unboxing null causes NullPointerException [OK]

Hint: Unboxing null causes runtime NullPointerException [OK]

Common Mistakes:

• Expecting compilation error instead of runtime error
• Assuming null unboxes to zero
• Thinking it prints 'null' string

5. Given this code, what will be the output?

List<Integer> list = List.of(1, 2, null, 4);
int sum = 0;
for (Integer n : list) {
    sum += n; // unboxing happens here
}
System.out.println(sum);

hard

A. Runtime NullPointerException at unboxing null

B. Compilation error due to null in list

C. Sum ignoring null, prints 7

D. 10

Solution

1. Step 1: Understand unboxing in loop with null element

   List contains a null Integer; unboxing null causes error.

2. Step 2: Identify runtime behavior

   When loop reaches null, unboxing triggers NullPointerException.

3. Final Answer:

   Runtime NullPointerException at unboxing null -> Option A

4. Quick Check:

   Unboxing null in loop causes runtime exception [OK]

Hint: Unboxing null in collections causes runtime exception [OK]

Common Mistakes:

• Assuming null is skipped or treated as zero
• Expecting compilation error due to null in list
• Thinking sum is partial ignoring null