Primitive vs reference storage in Java - Practice Questions

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of primitive and reference variable changes

What is the output of the following Java code?

Java

public class Test {
    public static void main(String[] args) {
        int a = 5;
        int b = a;
        b = 10;
        System.out.println("a = " + a);
        System.out.println("b = " + b);
    }
}

A a = 5\nb = 10

B a = 5\nb = 5

C a = 10\nb = 10

D a = 10\nb = 5

💡 Hint

Remember that primitives store values directly, so changing b does not affect a.

In Java, primitive types like int store values directly. When b = a, b gets a copy of the value 5. Changing b to 10 does not change a, so a remains 5.

❓ Predict Output

intermediate

Output when modifying an object reference

What is the output of this Java code?

Java

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

A 1

B 10

C Compilation error

D 0

💡 Hint

Arrays are objects, so arr1 and arr2 point to the same data.

In Java, arrays are reference types. Both arr1 and arr2 point to the same array. Changing arr2[0] changes the first element of the array, so arr1[0] is also 10.

❓ Predict Output

advanced

Effect of reassigning object references

What is the output of this Java code?

Java

public class Test {
    public static void main(String[] args) {
        StringBuilder sb1 = new StringBuilder("Hello");
        StringBuilder sb2 = sb1;
        sb2.append(" World");
        sb2 = new StringBuilder("New");
        System.out.println(sb1.toString());
    }
}

A New

B Hello

C Hello World

D Compilation error

💡 Hint

Appending changes the object both references point to; reassigning sb2 does not affect sb1.

sb1 and sb2 initially point to the same StringBuilder object. Appending " World" modifies that object. Then sb2 is reassigned to a new object, but sb1 still points to the original modified object. So sb1.toString() prints "Hello World".

❓ Predict Output

advanced

Output when passing primitives and objects to methods

What is the output of this Java code?

Java

public class Test {
    public static void main(String[] args) {
        int x = 5;
        StringBuilder sb = new StringBuilder("Hi");
        modify(x, sb);
        System.out.println(x + " " + sb.toString());
    }
    static void modify(int a, StringBuilder b) {
        a = 10;
        b.append(" there");
    }
}

A 5 Hi

B 10 Hi there

C 10 Hi

D 5 Hi there

💡 Hint

Primitives are passed by value; objects references allow modification of the object.

The primitive x is passed by value, so changing a inside modify does not affect x. The StringBuilder sb reference allows modification of the object, so " there" is appended. The output is "5 Hi there".

🧠 Conceptual

expert

Understanding memory behavior of primitives and references

Which statement best describes how Java stores primitive types and reference types in memory?

A Primitive variables store the actual data value; reference variables store the memory address of the object.

B Primitive variables store the memory address of the object; reference variables store the actual data value.

C Both primitive and reference variables store actual data values directly in the stack memory.

D Both primitive and reference variables store memory addresses pointing to objects in the heap.

💡 Hint

Think about where the actual data and references are stored for primitives and objects.

In Java, primitive variables hold the actual data value directly (e.g., int holds the number). Reference variables hold the address (reference) to the object stored in the heap memory.

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