Value equality in records in C Sharp (C#) - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a record named Person with two properties.

C Sharp (C#)

public record Person(string [1], int Age);

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

AName

BfirstName

CFullName

Dname

Attempts:

3 left

💡 Hint

Common Mistakes

Using lowercase property names which is against C# naming conventions.

Using unrelated property names.

The record property should be named Name to match common conventions.

2fill in blank

medium

Complete the code to create two Person records with the same values.

C Sharp (C#)

var person1 = new Person("Alice", 30);
var person2 = new Person("Alice", [1]);

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

A25

B30

C35

D40

Attempts:

3 left

💡 Hint

Common Mistakes

Using different ages which makes the records unequal.

Confusing the age values.

Both records must have the same age 30 to be equal by value.

3fill in blank

hard

Fix the error in the equality check between two Person records.

C Sharp (C#)

bool areEqual = person1 [1] person2;

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

A==

B=

C!=

Dequals

Attempts:

3 left

💡 Hint

Common Mistakes

Using single = which is assignment, not comparison.

Using != which checks inequality.

Using 'equals' which is not valid syntax in C#.

Use the == operator to check value equality between records.

4fill in blank

hard

Fill both blanks to override the ToString method in the Person record.

C Sharp (C#)

public override string ToString() => $"[1]: [2]";

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

AName

BAge

CName + Age

DPerson

Attempts:

3 left

💡 Hint

Common Mistakes

Using combined strings instead of separate properties.

Using class name instead of property names.

The ToString method should return the person's Name and Age values.

5fill in blank

hard

Fill all three blanks to create a new Person record with updated Age using the with-expression.

C Sharp (C#)

var olderPerson = person1 with { [1] = [2] + [3] };

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

AAge

Bperson1.Age

C1

DName

Attempts:

3 left

💡 Hint

Common Mistakes

Trying to update the Name property instead of Age.

Using incorrect syntax for the with-expression.

The with-expression updates the Age property by adding 1 to person1.Age.

Practice

1. What is the main feature of record types in C# regarding equality?

easy

A. They require manual implementation of equality methods.

B. They compare objects by their memory address only.

C. They cannot be compared for equality.

D. They compare objects by their data automatically.

Solution

1. Step 1: Understand record equality behavior

   Records in C# automatically implement value equality, meaning two records with the same data are considered equal.

2. Step 2: Contrast with normal classes

   Normal classes compare by reference (memory address) unless equality is overridden manually.

3. Final Answer:

   They compare objects by their data automatically. -> Option D

4. Quick Check:

   Record equality = value-based [OK]

Hint: Records compare data, not references, by default [OK]

Common Mistakes:

• Thinking records compare by reference like classes
• Assuming manual equality override is needed
• Believing records cannot be compared

2. Which of the following is the correct syntax to declare a positional record in C#?

easy

A. class Person { public string Name; }

B. record Person { string Name; }

C. record Person(string Name);

D. struct Person(string Name);

Solution

1. Step 1: Identify record declaration syntax

   The concise syntax for a record with a positional parameter is record Person(string Name);.

2. Step 2: Check other options

   class Person { public string Name; } is a class, not a record. struct Person(string Name); is a struct, not a record. record Person { string Name; } is invalid because positional parameters require parentheses.

3. Final Answer:

   record Person(string Name); -> Option C

4. Quick Check:

   Record syntax = record Name(params) [OK]

Hint: Use 'record Name(params);' for simple record declaration [OK]

Common Mistakes:

• Using class instead of record
• Missing parentheses for positional parameters
• Confusing struct with record

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

record Point(int X, int Y);

var p1 = new Point(3, 4);
var p2 = new Point(3, 4);
Console.WriteLine(p1 == p2);

medium

A. True

B. Compilation error

C. False

D. Runtime exception

Solution

1. Step 1: Understand record equality operator

   Records override the equality operator to compare values, so p1 == p2 compares the data inside the records.

2. Step 2: Compare values of p1 and p2

   Both have X=3 and Y=4, so they are equal and the output is True.

3. Final Answer:

   True -> Option A

4. Quick Check:

   Records with same data == True [OK]

Hint: Records compare data with == operator automatically [OK]

Common Mistakes:

• Assuming == compares references for records
• Expecting compilation error due to operator
• Thinking output is False because objects differ

4. Identify the error in this code snippet:

record Person(string Name);

var p1 = new Person("Alice");
var p2 = new Person("Alice");

if (p1.Equals(p2))
    Console.WriteLine("Equal");
else
    Console.WriteLine("Not Equal");

p1.Name = "Bob";

medium

A. Records cannot use Equals method.

B. Cannot assign to 'Name' because it is init-only.

C. Missing semicolon after record declaration.

D. p1 and p2 are different objects, so Equals returns false.

Solution

1. Step 1: Check mutability of record properties

   By default, positional record properties are immutable (init-only), so p1.Name = "Bob"; causes a compile error.

2. Step 2: Verify other statements

   Using Equals is valid and compares values correctly. The semicolon after record declaration is present. Equals returns true because data matches.

3. Final Answer:

   Cannot assign to 'Name' because it is init-only. -> Option B

4. Quick Check:

   Record properties are immutable by default [OK]

Hint: Record properties are init-only; cannot assign after creation [OK]

Common Mistakes:

• Trying to modify record properties after creation
• Thinking Equals compares references
• Ignoring immutability of positional properties

5. Given two records record Product(int Id, string Name);, how can you create a new record from an existing one but with a different Name while keeping Id the same?

hard

A. Use the 'with' expression: var p2 = p1 with { Name = "NewName" };

B. Create a new record manually: var p2 = new Product(p1.Name, "NewName");

C. Assign directly: p1.Name = "NewName";

D. Use a method to clone and change: var p2 = p1.Clone("NewName");

Solution

1. Step 1: Understand the 'with' expression for records

   The 'with' expression creates a copy of a record with specified properties changed.

2. Step 2: Compare options

   Use the 'with' expression: var p2 = p1 with { Name = "NewName" }; uses 'with' correctly. Create a new record manually: var p2 = new Product(p1.Name, "NewName"); fails to compile because Id expects int but p1.Name is string. Assign directly: p1.Name = "NewName"; is invalid because record properties are immutable. Use a method to clone and change: var p2 = p1.Clone("NewName"); is invalid as no Clone method exists by default.

3. Final Answer:

   Use the 'with' expression: var p2 = p1 with { Name = "NewName" }; -> Option A

4. Quick Check:

   'with' creates modified record copies [OK]

Hint: Use 'with' to copy and change record properties easily [OK]

Common Mistakes:

• Trying to assign directly to immutable properties
• Not using 'with' for copying records
• Assuming a Clone method exists by default