Record structs in C Sharp (C#) - Step-by-Step Execution

Concept Flow - Record structs

Define record struct

↓

Create instance with values

↓

Use instance: access properties

↓

Compare instances (value equality)

↓

Immutable by default, can use with-expressions

↓

End

This flow shows how a record struct is defined, instantiated, used, compared, and how its immutability works.

Execution Sample

C Sharp (C#)

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

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

Console.WriteLine(p1 == p2);

Defines a record struct Point, creates two instances with same values, and compares them for equality.

Execution Table

Step	Action	Evaluation	Result
1	Define record struct Point with properties X and Y	N/A	Point type created
2	Create p1 = new Point(3, 4)	X=3, Y=4	p1 holds (3,4)
3	Create p2 = new Point(3, 4)	X=3, Y=4	p2 holds (3,4)
4	Compare p1 == p2	Compare values of X and Y	True
5	Print result	Output boolean	True

💡 Comparison done, program ends after printing True

Variable Tracker

Variable	Start	After Step 2	After Step 3	Final
p1	undefined	(3,4)	(3,4)	(3,4)
p2	undefined	undefined	(3,4)	(3,4)

Key Moments - 3 Insights

Why does p1 == p2 return True even though they are different instances?

Record structs compare their contents (values) for equality, not their memory addresses. See execution_table step 4 where values (3,4) are compared.

Are record structs mutable like classes?

No, record structs are immutable by default, meaning their properties cannot be changed after creation unless explicitly declared mutable.

What happens if we try to change a property of p1 after creation?

Since properties are readonly by default, the compiler will give an error. You must create a new instance with desired changes.

Visual Quiz - 3 Questions

Look at the execution_table at step 4, what is the result of comparing p1 and p2?

A False

B True

C Compilation error

D Runtime exception

💡 Hint

Check the 'Result' column in execution_table row 4.

According to variable_tracker, what is the value of p2 after step 3?

A (0,0)

B undefined

C (3,4)

D null

💡 Hint

Look at variable_tracker row for p2 under 'After Step 3'.

If we try to assign p1.X = 5 after creation, what will happen?

A Compilation error due to immutability

B Runtime exception

C The value of p1.X changes to 5

D p1 becomes null

💡 Hint

Refer to key_moments about immutability and property changes.

Concept Snapshot

Record structs are value types with immutable properties by default.
They support value-based equality (== compares contents).
Defined with 'record struct' syntax.
Instances are created with positional parameters.
Properties cannot be changed after creation unless mutable.
Use with-expressions to create modified copies.

Full Transcript

This example shows how to define a record struct named Point with two properties X and Y. We create two instances p1 and p2 with the same values (3,4). When we compare p1 and p2 using ==, the result is True because record structs compare their values, not references. The variables p1 and p2 hold the values (3,4) after creation. Record structs are immutable by default, so trying to change a property after creation causes a compilation error. This behavior ensures safe and predictable value semantics.

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