Unboxing in Java - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to unbox the Integer object to an int primitive.

Java

Integer num = 10;
int value = num[1];

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

A.toString()

B.valueOf()

C.intValue()

D.doubleValue()

Attempts:

3 left

💡 Hint

Common Mistakes

Using toString() returns a String, not an int.

Using valueOf() is a static method to box, not unbox.

The method intValue() converts the Integer object to a primitive int.

2fill in blank

medium

Complete the code to assign the unboxed Double value to a double variable.

Java

Double d = 5.5;
double val = d[1];

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

A.toString()

B.intValue()

C.floatValue()

D.doubleValue()

Attempts:

3 left

💡 Hint

Common Mistakes

Using intValue() loses decimal precision.

Using toString() returns a String, not a double.

doubleValue() returns the primitive double from a Double object.

3fill in blank

hard

Fix the error in unboxing the Boolean object to a boolean primitive.

Java

Boolean flag = Boolean.TRUE;
boolean isTrue = flag[1];

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

A.booleanValue()

B.intValue()

C.toString()

D.doubleValue()

Attempts:

3 left

💡 Hint

Common Mistakes

Using intValue() causes a compile error because Boolean has no such method.

Using toString() returns a String, not a boolean.

booleanValue() returns the primitive boolean from a Boolean object.

4fill in blank

hard

Fill both blanks to unbox the Character object and compare it to a char literal.

Java

Character ch = 'a';
if (ch[1] == '[2]') {
    System.out.println("Match");
}

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

A.charValue()

Ba

Cb

D.toString()

Attempts:

3 left

💡 Hint

Common Mistakes

Using toString() returns a String, which cannot be compared to a char literal.

Comparing to the wrong char literal causes the condition to fail.

charValue() unboxes Character to char, which can be compared to a char literal 'a'.

5fill in blank

hard

Fill all three blanks to create a map with unboxed keys and values from wrapper objects.

Java

Map<Integer, Double> map = new HashMap<>();
Integer key = 1;
Double value = 2.5;
map.put(key[1], value[2]);

for (Map.Entry<Integer, Double> entry : map.entrySet()) {
    int k = entry.getKey();
    double v = entry.getValue();
    System.out.println(k [3] v);
}

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

A.intValue()

B.doubleValue()

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Not unboxing before putting into the map uses the wrapper objects directly.

Using - instead of + changes the output meaning.

Forgetting to unbox compiles but does not demonstrate explicit unboxing.

intValue() and doubleValue() unbox the wrapper objects; + adds the values in the output.

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