Record structs in C Sharp (C#) - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of record struct with property init

What is the output of this C# code snippet using a record struct?

C Sharp (C#)

public record struct Point(int X, int Y);

var p1 = new Point(3, 4);
var p2 = p1 with { Y = 10 };
Console.WriteLine(p2);

A Point { X = 3, Y = 4 }

B Point { X = 3, Y = 10 }

C Compilation error: 'with' expressions not supported on record structs

D Point { X = 10, Y = 4 }

💡 Hint

Remember that record structs support 'with' expressions to create copies with modifications.

The 'with' expression creates a copy of the record struct with the specified property changed. Here, Y is changed to 10 while X remains 3.

🧠 Conceptual

intermediate

Immutability of record structs

Which statement about record structs in C# is true?

A Record structs are always immutable and cannot have mutable properties.

B Record structs behave exactly like classes in terms of reference equality.

C Record structs can have mutable properties but still provide value equality semantics.

D Record structs cannot have parameterized constructors.

💡 Hint

Think about how record structs differ from classes and normal structs.

Record structs provide value equality and can have mutable properties. They are value types, so they differ from classes in reference behavior.

🔧 Debug

advanced

Fix the error in record struct declaration

This code produces a compilation error. What is the cause?

C Sharp (C#)

public record struct Rectangle
{
    public int Width { get; init; }
    public int Height { get; init; }

    public Rectangle(int width, int height) : this()
    {
        Width = width;
        Height = height;
    }
}

A The constructor must use 'this()' to call the parameterless constructor before assigning properties.

B Record structs cannot have properties with 'init' accessors.

C The properties must be declared as 'readonly' to be used in a record struct.

D The constructor must be removed because record structs cannot have explicit constructors.

💡 Hint

In structs, constructors must initialize all fields and call the parameterless constructor if present.

In C#, struct constructors must call the parameterless constructor (or chain to another constructor) before assigning to properties. Omitting 'this()' causes a compilation error.

📝 Syntax

advanced

Identify the invalid record struct syntax

Which of the following record struct declarations is invalid in C#?

A public record struct Data(int X, int Y) : BaseStruct;

B public record struct Car { public string Model { get; init; } public int Year { get; init; } }

C public record struct Person(string Name, int Age);

D public record struct Point { public int X; public int Y; }

💡 Hint

Consider inheritance rules for structs in C#.

Structs, including record structs, cannot inherit from other structs or classes, so ': BaseStruct' is invalid syntax.

🚀 Application

expert

Value equality behavior of record structs

Given this code, what is the output?

C Sharp (C#)

public record struct Coordinate(int X, int Y);

var c1 = new Coordinate(5, 7);
var c2 = new Coordinate(5, 7);

Console.WriteLine(c1 == c2);
Console.WriteLine(ReferenceEquals(c1, c2));

A False\nTrue

B False\nFalse

C True\nTrue

D True\nFalse

💡 Hint

Record structs provide value equality but are value types, so reference equality behaves differently.

The '==' operator for record structs compares values and returns true. ReferenceEquals returns false because structs are value types and boxed separately.

Practice

1. What is a key feature of record struct in C#?

easy

A. It provides value-based equality and immutability by default.

B. It allows inheritance from other structs.

C. It can only be used for reference types.

D. It does not support property declarations.

Solution

1. Step 1: Understand record struct purpose

   Record structs combine struct's lightweight nature with record's value-based equality and immutability.

2. Step 2: Analyze options

   It provides value-based equality and immutability by default. correctly states value-based equality and immutability. Options B, C, and D are incorrect because structs cannot inherit, record structs are value types, and they support properties.

3. Final Answer:

   It provides value-based equality and immutability by default. -> Option A

4. Quick Check:

   record struct = value equality + immutability [OK]

Hint: Remember: record structs behave like immutable value containers [OK]

Common Mistakes:

• Thinking record structs support inheritance
• Confusing record structs with classes
• Assuming record structs are reference types

2. Which of the following is the correct syntax to declare a record struct named Point with two properties X and Y of type int?

easy

A. struct record Point { int X; int Y; }

B. record Point struct { int X; int Y; }

C. record struct Point(int X, int Y);

D. record struct Point { int X; int Y; }

Solution

1. Step 1: Recall record struct declaration syntax

   The concise syntax for record structs uses positional parameters: record struct Name(type prop1, type prop2);

2. Step 2: Compare options

   record struct Point(int X, int Y); matches the correct syntax. record struct Point { int X; int Y; } is close but missing parentheses for positional parameters. Options A and C have incorrect order or keywords.

3. Final Answer:

   record struct Point(int X, int Y); -> Option C

4. Quick Check:

   record struct with positional params = record struct Point(int X, int Y); [OK]

Hint: Use parentheses for positional properties in record structs [OK]

Common Mistakes:

• Swapping 'record' and 'struct' keywords
• Omitting parentheses for positional parameters
• Using braces instead of parentheses for positional syntax

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

record struct 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. False

C. Compilation error

D. Runtime exception

Solution

1. Step 1: Understand equality in record structs

   Record structs provide value-based equality by default, so two instances with same values are equal.

2. Step 2: Analyze the code output

   Both p1 and p2 have X=3 and Y=4, so p1 == p2 evaluates to true.

3. Final Answer:

   True -> Option A

4. Quick Check:

   record struct equality = value equality [OK]

Hint: Record structs compare values, not references [OK]

Common Mistakes:

• Assuming == compares references for structs
• Expecting compilation error for == operator
• Confusing record structs with classes for equality

4. Identify the error in this code snippet:

record struct Rectangle(int Width, int Height);
var rect = new Rectangle(5, 10);
rect.Width = 7;
Console.WriteLine(rect);

medium

A. Rectangle must be a class to allow property assignment.

B. Missing semicolon after record struct declaration.

C. Console.WriteLine cannot print record structs.

D. Cannot assign to 'Width' because it is readonly.

Solution

1. Step 1: Understand immutability of record structs

   Record structs with positional parameters generate readonly properties by default.

2. Step 2: Analyze the assignment line

   Trying to assign rect.Width = 7; causes a compile-time error because Width is readonly.

3. Final Answer:

   Cannot assign to 'Width' because it is readonly. -> Option D

4. Quick Check:

   Positional record struct properties are readonly [OK]

Hint: Positional record struct properties are readonly by default [OK]

Common Mistakes:

• Trying to modify readonly properties
• Confusing record structs with mutable classes
• Ignoring compiler error messages

5. You want to create a lightweight data type to store a 3D point with coordinates X, Y, and Z. You want it to be immutable, support value equality, and print nicely. Which is the best way to declare it using record structs?

hard

A. struct Point3D { public int X; public int Y; public int Z; }

B. record struct Point3D(int X, int Y, int Z);

C. class Point3D { public int X { get; set; } public int Y { get; set; } public int Z { get; set; } }

D. record class Point3D(int X, int Y, int Z);

Solution

1. Step 1: Identify requirements

   Needs lightweight, immutable, value equality, and nice printing.

2. Step 2: Match requirements to options

   record struct Point3D(int X, int Y, int Z); uses record struct with positional parameters, providing all requirements. struct Point3D { public int X; public int Y; public int Z; } is mutable struct without value equality or nice printing. class Point3D { public int X { get; set; } public int Y { get; set; } public int Z { get; set; } } is mutable class, heavier and reference type. record class Point3D(int X, int Y, int Z); is record class, reference type, not lightweight.

3. Final Answer:

   record struct Point3D(int X, int Y, int Z); -> Option B

4. Quick Check:

   record struct = lightweight + immutable + value equality [OK]

Hint: Use record struct with positional params for immutable lightweight data [OK]

Common Mistakes:

• Choosing class instead of struct for lightweight data
• Using mutable fields instead of readonly properties
• Confusing record class with record struct