Array as reference type behavior in C Sharp (C#) - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare an integer array of size 3.

C Sharp (C#)

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

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

A1

B5

C0

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Using zero or negative numbers as array size causes errors.

Forgetting to specify the size when creating the array.

The array size is specified inside the brackets when creating a new array. Here, the size is 3.

2fill in blank

medium

Complete the code to assign the value 10 to the first element of the array.

C Sharp (C#)

numbers[[1]] = 10;

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

A1

B0

C2

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 as the first index, which causes off-by-one errors.

Using an index outside the array bounds.

Array indexes in C# start at 0, so the first element is at index 0.

3fill in blank

hard

Fix the error in the code to copy the reference of the array correctly.

C Sharp (C#)

int[] copy = [1];

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

Anew int[3]

Bnumbers.Clone()

Cnumbers

Dnumbers.CopyTo()

Attempts:

3 left

💡 Hint

Common Mistakes

Using Clone() creates a shallow copy, not just a reference copy.

Using CopyTo() requires parameters and does not return an array.

Assigning one array variable to another copies the reference, so both point to the same array.

4fill in blank

hard

Fill both blanks to show that changing the copy affects the original array.

C Sharp (C#)

copy[[1]] = 20;
Console.WriteLine(numbers[[2]]);

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

A1

B0

C2

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Using different indexes for copy and numbers, which shows unrelated elements.

Using an index outside the array bounds.

Since both variables reference the same array, changing one affects the other at the same index.

5fill in blank

hard

Fill all three blanks to create a new array copy and change it without affecting the original.

C Sharp (C#)

int[] newCopy = new int[numbers.Length];
Array.[1](numbers, newCopy, numbers.[2]);
newCopy[[3]] = 30;
Console.WriteLine(numbers[1]);

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

ACopy

BLength

C1

DClone

Attempts:

3 left

💡 Hint

Common Mistakes

Using Clone creates a shallow copy but returns an object, not an array.

Changing the original array instead of the copy.

Using Array.Copy copies elements to a new array. Changing newCopy does not affect numbers.

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