Array as reference type behavior in C Sharp (C#) - Mini Project: Build & Apply

Array as reference type behavior

📖 Scenario: Imagine you have a list of favorite fruits stored in an array. You want to see how changing this list in one place affects the list in another place because arrays in C# work as reference types.

🎯 Goal: You will create an array of fruits, assign it to another variable, change one fruit through the new variable, and observe how the original array changes too.

📋 What You'll Learn

Create an array called fruits with these exact values: "Apple", "Banana", "Cherry"

Create a new array variable called favoriteFruits and assign it the value of fruits

Change the second element (index 1) of favoriteFruits to "Blueberry"

Print the contents of the fruits array to show the change

💡 Why This Matters

🌍 Real World

Understanding how arrays work as reference types helps when managing lists of data like user inputs, settings, or items in a shopping cart.

💼 Career

Many programming jobs require managing collections of data efficiently and avoiding bugs caused by unintended changes to shared data.

1

Create the initial array

Create an array called fruits with these exact values: "Apple", "Banana", "Cherry"

C Sharp (C#)

# Create an array called fruits with "Apple", "Banana", "Cherry"
// Your code here

Hint

Use string[] fruits = { "Apple", "Banana", "Cherry" }; to create the array.

2

Assign the array to a new variable

Create a new array variable called favoriteFruits and assign it the value of fruits

C Sharp (C#)

string[] fruits = { "Apple", "Banana", "Cherry" };
// Assign fruits to favoriteFruits
// Your code here

Hint

Use string[] favoriteFruits = fruits; to assign the array reference.

3

Change an element through the new variable

Change the second element (index 1) of favoriteFruits to "Blueberry"

C Sharp (C#)

string[] fruits = { "Apple", "Banana", "Cherry" };
string[] favoriteFruits = fruits;
// Change favoriteFruits[1] to "Blueberry"
// Your code here

Hint

Use favoriteFruits[1] = "Blueberry"; to change the second element.

4

Print the original array to see the change

Use a for loop with variable i from 0 to fruits.Length to print each element of the fruits array on its own line

C Sharp (C#)

string[] fruits = { "Apple", "Banana", "Cherry" };
string[] favoriteFruits = fruits;
favoriteFruits[1] = "Blueberry";
// Print each element of fruits using a for loop
// Your code here

Hint

Use a for loop from 0 to fruits.Length and print each element with Console.WriteLine(fruits[i]);.

Practice

1. In C#, what happens when you assign one array variable to another, like int[] a = {1, 2}; int[] b = a;?

easy

A. Both variables refer to the same array in memory.

B. A new copy of the array is created and assigned to b.

C. The array elements are copied individually to b.

D. The assignment causes a compile-time error.

Solution

1. Step 1: Understand array assignment in C#

   Arrays are reference types, so assigning one array variable to another copies the reference, not the data.

2. Step 2: Effect of assignment on variables

   Both variables point to the same array object in memory, so changes via one affect the other.

3. Final Answer:

   Both variables refer to the same array in memory. -> Option A

4. Quick Check:

   Array assignment copies reference, not data [OK]

Hint: Remember: arrays are references, assignment copies reference [OK]

Common Mistakes:

• Thinking assignment copies array elements
• Assuming a new array is created automatically
• Confusing arrays with value types

2. Which of the following is the correct syntax to declare and assign an integer array with values 5, 10, 15 in C#?

easy

A. int arr = {5, 10, 15};

B. int[] arr = new int[3] {5, 10, 15};

C. int[] arr = new int {5, 10, 15};

D. int arr[] = new int[] {5, 10, 15};

Solution

1. Step 1: Review array declaration syntax

   In C#, arrays are declared with the type followed by square brackets, and initialized with new and size or values.

2. Step 2: Check each option

   int[] arr = new int[3] {5, 10, 15}; uses correct syntax with size and initializer. int arr = {5, 10, 15}; misses array brackets. int[] arr = new int {5, 10, 15}; misses the square brackets after int. int arr[] = new int[] {5, 10, 15}; uses C-style syntax which is invalid in C#.

3. Final Answer:

   int[] arr = new int[3] {5, 10, 15}; -> Option B

4. Quick Check:

   Correct array declaration uses new int[size] {values} [OK]

Hint: Use int[] arr = new int[size] {values}; for arrays [OK]

Common Mistakes:

• Omitting square brackets in declaration
• Using C-style array syntax
• Missing new keyword with size

3. What is the output of the following C# code?

int[] a = {1, 2, 3};
int[] b = a;
b[0] = 10;
Console.WriteLine(a[0]);

medium

A. Compilation error

B. 1

C. 0

D. 10

Solution

1. Step 1: Understand reference assignment

   Variable b is assigned the reference of array a, so both point to the same array.

2. Step 2: Effect of modifying b[0]

   Changing b[0] to 10 changes the first element of the shared array.

3. Step 3: Output of a[0]

   Since a and b reference the same array, a[0] is now 10.

4. Final Answer:

   10 -> Option D

5. Quick Check:

   Arrays share reference, so change reflects in both [OK]

Hint: Changing one reference changes the shared array [OK]

Common Mistakes:

• Expecting original array to remain unchanged
• Confusing value copy with reference copy
• Assuming separate arrays after assignment

4. Consider this code snippet:

int[] x = {4, 5, 6};
int[] y = new int[3];
y = x;
y[1] = 20;
Console.WriteLine(x[1]);

What is the output and why?

medium

A. 5, because y was a new array and changes don't affect x.

B. 0, because y was initialized with zeros.

C. 20, because y references the same array as x after assignment.

D. Compilation error due to invalid assignment.

Solution

1. Step 1: Analyze array assignment

   Initially, y points to a new array, but then y = x; makes y reference the same array as x.

2. Step 2: Effect of modifying y[1]

   Changing y[1] to 20 modifies the shared array, so x[1] also becomes 20.

3. Final Answer:

   20, because y references the same array as x after assignment. -> Option C

4. Quick Check:

   Reference assignment means both variables share array [OK]

Hint: Assignment copies reference, so changes affect original array [OK]

Common Mistakes:

• Thinking y remains a separate array after assignment
• Assuming array elements are copied automatically
• Confusing initialization with assignment

5. You want to create a copy of an integer array original so that modifying the copy does NOT affect the original. Which code correctly achieves this in C#?

hard

A. Both B and D

B. int[] copy = (int[])original.Clone();

C. int[] copy = original;

D. int[] copy = original.Select(x => x).ToArray();

Solution

1. Step 1: Understand reference vs copy

   Assigning copy = original; copies the reference, so changes affect both.

2. Step 2: Using Clone() and LINQ Select

   Clone() creates a shallow copy of the array, and Select with ToArray() creates a new array with copied elements.

3. Step 3: Confirm both methods create independent arrays

   Both Clone() and Select(...).ToArray() produce new arrays, so modifying the copy won't affect the original.

4. Final Answer:

   Both B and D -> Option A

5. Quick Check:

   Use Clone() or LINQ to copy arrays [OK]

Hint: Use Clone() or LINQ to copy arrays, not assignment [OK]

Common Mistakes:

• Using assignment instead of copying
• Assuming Clone() returns reference
• Not realizing LINQ creates new array