Array as reference type behavior in C Sharp (C#) - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of modifying array through reference

What is the output of the following C# code?

C Sharp (C#)

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

A 0

B 1

C Compilation error

D 10

💡 Hint

Arrays in C# are reference types, so changes via one reference affect the original array.

Both 'arr' and 'refArr' point to the same array in memory. Changing refArr[0] changes arr[0].

❓ Predict Output

intermediate

Effect of reassigning array reference

What is the output of this C# code snippet?

C Sharp (C#)

int[] original = {5, 6, 7};
int[] copy = original;
copy = new int[] {1, 2, 3};
Console.WriteLine(original[0]);

A 5

B 1

C 0

D Runtime error

💡 Hint

Reassigning 'copy' does not change 'original'.

'copy' initially points to 'original' array, but after reassignment it points to a new array. 'original' remains unchanged.

🔧 Debug

advanced

Why does this array modification not affect original?

Consider this code: int[] a = {1, 2, 3}; ModifyArray(a); Console.WriteLine(a[0]); void ModifyArray(int[] arr) { arr = new int[] {9, 8, 7}; } Why does the output remain 1?

C Sharp (C#)

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

void ModifyArray(int[] arr) {
    arr = new int[] {9, 8, 7};
}

A Because arrays are value types and passed by value

B Because the new array is assigned but not returned

C Because the parameter 'arr' is a copy of the reference, reassignment doesn't affect original

D Because the method does not modify any element of the original array

💡 Hint

Think about how references are passed to methods in C#.

The method parameter 'arr' is a copy of the reference. Reassigning it points it to a new array, but does not change the caller's reference.

❓ Predict Output

advanced

Output after modifying array element inside method

What is the output of this C# code?

C Sharp (C#)

int[] nums = {4, 5, 6};
ChangeFirst(nums);
Console.WriteLine(nums[0]);

void ChangeFirst(int[] arr) {
    arr[0] = 100;
}

A 100

B Compilation error

C 0

D 4

💡 Hint

Modifying elements of the array affects the original array.

The method changes the first element of the array that both 'nums' and 'arr' reference.

🧠 Conceptual

expert

Understanding array reference behavior with multiple assignments

Given the code below, what is the final output?

C Sharp (C#)

int[] x = {1, 2};
int[] y = x;
x[0] = 10;
y = new int[] {3, 4};
x[1] = 20;
Console.WriteLine(string.Join(",", y));

A 10,20

B 3,4

C 1,2

D 10,2

💡 Hint

Remember that 'y' is reassigned to a new array after 'x' is modified.

'y' initially points to the same array as 'x'. After reassignment, 'y' points to a new array {3,4}. The Console.WriteLine prints the contents of 'y', which is now {3,4}.

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