Primitive vs reference storage in Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a primitive integer variable.

Java

int number = [1];

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

A10

B"10"

Cnew Integer(10)

DInteger.valueOf(10)

Attempts:

3 left

💡 Hint

Common Mistakes

Using quotes around the number makes it a String, not an int.

Using 'new Integer(10)' creates an object, not a primitive.

The primitive type int stores the value directly. So assigning 10 is correct.

2fill in blank

medium

Complete the code to create a reference to a String object.

Java

String text = [1];

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

A10

Bnew int(5)

C"Hello"

Dtrue

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning a number or boolean to a String variable causes errors.

Using 'new int(5)' is invalid syntax and wrong type.

Strings are objects in Java, so they are referenced by variables. Use quotes to create a String literal.

3fill in blank

hard

Fix the error in the code to assign a new Integer object to a variable.

Java

Integer value = [1];

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

Anew Integer(10)

B"10"

C10

DInteger.parseInt("10")

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning a primitive int directly to an Integer variable without autoboxing.

Using a string literal without conversion causes errors.

To assign an Integer object explicitly, use new Integer(10). The others are either primitives or methods returning int.

4fill in blank

hard

Fill both blanks to create an array of primitive ints and assign a value.

Java

int[] numbers = new int[[1]];
numbers[[2]] = 5;

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

A3

B0

C1

D5

Attempts:

3 left

💡 Hint

Common Mistakes

Using an index equal to the size causes an error.

Using negative indices is invalid.

The array size is 3, and the first index is 0. So assigning 5 to numbers[0] is correct.

5fill in blank

hard

Fill all three blanks to create a String array, assign a value, and print it.

Java

String[] words = new String[[1]];
words[[2]] = [3];
System.out.println(words[0]);

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

A2

B0

C"hello"

D"world"

Attempts:

3 left

💡 Hint

Common Mistakes

Using an index outside the array size causes errors.

Assigning strings without quotes causes errors.

The array size is 2, index 0 is assigned the string "hello", and printing words[0] outputs 'hello'.

Practice

1. Which of the following correctly describes how primitive types are stored in Java?

easy

A. They store values as strings internally.

B. They store the actual value directly in the variable.

C. They store the memory address of the variable.

D. They store a reference to an object in memory.

Solution

1. Step 1: Understand primitive storage

   Primitive types like int, double, and boolean store the actual data value directly in the variable's memory location.

2. Step 2: Compare with reference types

   Reference types store addresses pointing to objects, not the actual data value.

3. Final Answer:

   They store the actual value directly in the variable. -> Option B

4. Quick Check:

   Primitive = actual value [OK]

Hint: Primitives hold values; references hold addresses [OK]

Common Mistakes:

• Thinking primitives store memory addresses
• Confusing reference storage with primitive storage
• Assuming primitives store objects

2. Which of the following is the correct way to declare a reference variable for a String in Java?

easy

A. int s = new String();

B. boolean s = new String();

C. String s = 10;

D. String s = new String();

Solution

1. Step 1: Check variable type and assignment

   String is a reference type, so it can be assigned with new String() to create a new object.

2. Step 2: Verify type compatibility

   int and boolean are primitive types and cannot be assigned a String object.

3. Final Answer:

   String s = new String(); -> Option D

4. Quick Check:

   Reference declaration matches object creation [OK]

Hint: Reference types use 'new' with matching type [OK]

Common Mistakes:

• Assigning objects to primitive types
• Using incompatible types in declaration
• Missing 'new' keyword for object creation

3. What will be the output of the following code?

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

medium

A. 5

B. 10

C. 0

D. Compilation error

Solution

1. Step 1: Understand primitive assignment

   Variable b copies the value of a (which is 5). Changing b to 10 does not affect a.

2. Step 2: Print value of a

   Since a was not changed, printing a outputs 5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Primitive copy keeps original value [OK]

Hint: Changing copied primitive doesn't affect original [OK]

Common Mistakes:

• Assuming a changes when b changes
• Confusing reference behavior with primitive
• Expecting compilation error

4. Consider the code below. What is the error and how to fix it?

String s1 = "hello";
String s2 = s1;
s2 = "world";
System.out.println(s1);

medium

A. Output is 'world' because s2 changes s1.

B. Error: Strings cannot be assigned like this.

C. No error; output is 'hello'.

D. Compilation error due to missing new keyword.

Solution

1. Step 1: Analyze reference assignment

   s2 points to the same String object as s1 initially, but assigning s2 = "world" makes s2 point to a new String object.

2. Step 2: Check effect on s1

   s1 still points to "hello" because Strings are immutable and s2 reassignment does not affect s1.

3. Final Answer:

   No error; output is 'hello'. -> Option C

4. Quick Check:

   Reference reassignment doesn't change original object [OK]

Hint: Reassigning reference doesn't change original object [OK]

Common Mistakes:

• Thinking s1 changes when s2 changes
• Expecting compilation error for String assignment
• Confusing mutable and immutable objects

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

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

Box b1 = new Box(5);
Box b2 = b1;
b2.value = 10;
System.out.println(b1.value);

hard

A. 10

B. 5

C. Compilation error

D. NullPointerException

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 through b2

   Changing b2.value to 10 changes the single Box object's value.

3. Step 3: Print b1.value

   Since b1 points to the same object, b1.value is now 10.

4. Final Answer:

   10 -> Option A

5. Quick Check:

   Reference change affects original object [OK]

Hint: Object references share same data; change affects all [OK]

Common Mistakes:

• Assuming b1.value stays 5 after b2 changes it
• Confusing primitive and reference behavior
• Expecting compilation or runtime errors