Custom LINQ extension methods let you add your own easy-to-use commands to work with lists and collections. This helps you write cleaner and simpler code.
You want to reuse a special way to filter or change a list in many places.
You need a new operation on collections that LINQ does not provide by default.
You want your code to read like natural language when working with data.
You want to keep your main code clean by moving complex logic into small methods.
You want to chain multiple operations on collections smoothly.
Syntax
C Sharp (C#)
publicstaticclass MyExtensions
{
publicstatic IEnumerable<T> MyMethod<T>(this IEnumerable<T> source)
{
// Your code here
yield return default(T);
}
}
The method must be inside a static class.
The first parameter uses this keyword to mark it as an extension method.
Examples
This method returns only even numbers from a list of integers.
C Sharp (C#)
publicstaticclass Extensions
{
publicstatic IEnumerable<int> EvenNumbers(this IEnumerable<int> numbers)
{
foreach (var n in numbers)
if (n % 2 == 0)
yield return n;
}
}
This method returns every other item from a collection.
C Sharp (C#)
publicstaticclass Extensions
{
publicstatic IEnumerable<T> TakeEveryOther<T>(this IEnumerable<T> source)
{
bool take = true;
foreach (var item in source)
{
if (take) yield return item;
take = !take;
}
}
}
Sample Program
This program defines a custom extension method EvenNumbers that filters even numbers from a list. It then prints each even number.
C Sharp (C#)
using System;
using System.Collections.Generic;
using System.Linq;
publicstaticclass Extensions
{
publicstatic IEnumerable<int> EvenNumbers(this IEnumerable<int> numbers)
{
foreach (var n in numbers)
if (n % 2 == 0)
yield return n;
}
}
class Program
{
staticvoid Main()
{
var nums = newList<int> {1, 2, 3, 4, 5, 6};
var evens = nums.EvenNumbers();
foreach (var e in evens)
Console.WriteLine(e);
}
}
OutputSuccess
Important Notes
Extension methods help keep your code clean and readable.
Use yield return to return items one by one without creating a new list.
Remember to include using for the namespace where your extension methods live.
Summary
Custom LINQ extension methods add new commands to collections.
They must be static methods inside static classes with this on the first parameter.
They make your code easier to read and reuse.
Practice
(1/5)
1. What is a key requirement for creating a custom LINQ extension method in C#?
easy
A. It must be a static method inside a static class with this on the first parameter.
B. It must be an instance method inside any class.
C. It must return void and take no parameters.
D. It must be declared inside the Main method.
Solution
Step 1: Understand extension method structure
Extension methods must be static methods inside static classes.
Step 2: Identify the special parameter usage
The first parameter must have the this keyword to specify the type it extends.
Final Answer:
It must be a static method inside a static class with this on the first parameter. -> Option A
Quick Check:
Extension method = static + static class + this param [OK]
Hint: Remember: static method + static class + this param [OK]
Common Mistakes:
Making the method non-static
Forgetting the static class container
Omitting the 'this' keyword on the first parameter
2. Which of the following is the correct syntax to declare a custom LINQ extension method named IsEven for IEnumerable<int>?
easy
A. public static bool IsEven(this IEnumerable<int> source) { ... }
B. public bool IsEven(IEnumerable<int> source) { ... }
C. static bool IsEven(IEnumerable<int> source) { ... }
D. public static bool IsEven(IEnumerable<int> source) { ... }
Solution
Step 1: Check method modifiers and parameters
The method must be public static and have this before the first parameter to be an extension method.
Step 2: Verify the parameter type and method signature
public static bool IsEven(this IEnumerable<int> source) { ... } correctly uses this IEnumerable<int> source and is static and public.
Final Answer:
public static bool IsEven(this IEnumerable<int> source) { ... } -> Option A
Quick Check:
Extension method syntax = public static + this param [OK]
Hint: Look for 'public static' and 'this' on first parameter [OK]
Common Mistakes:
Missing 'this' keyword on the first parameter
Not making the method static
Incorrect access modifier
3. Given this custom LINQ extension method, what will be the output?
public static IEnumerable<int> GetPositive(this IEnumerable<int> source) {
foreach (var num in source) {
if (num > 0) yield return num;
}
}
var numbers = new List<int> { -1, 0, 3, 5, -2 };
var result = numbers.GetPositive().ToList();
Console.WriteLine(string.Join(",", result));
medium
A. -1,0,3,5,-2
B. 0,3,5
C. 3,5,-2
D. 3,5
Solution
Step 1: Understand the extension method logic
The method yields only numbers greater than zero from the source list.
Step 2: Apply the method to the list
From the list [-1, 0, 3, 5, -2], only 3 and 5 are greater than zero and yielded.
Final Answer:
3,5 -> Option D
Quick Check:
Filter positive numbers = 3,5 [OK]
Hint: Focus on the 'if (num > 0)' filter condition [OK]
Common Mistakes:
Including zero as positive
Returning all numbers without filtering
Confusing yield return with return
4. Identify the error in this custom LINQ extension method:
public static IEnumerable<int> FilterEven(IEnumerable<int> source) {
foreach (var num in source) {
if (num % 2 == 0) yield return num;
}
}
medium
A. Cannot use 'yield return' inside a foreach loop.
B. Method should return void instead of IEnumerable<int>.
C. Missing 'this' keyword on the first parameter to make it an extension method.
D. The modulo operator '%' is not valid in C#.
Solution
Step 1: Check extension method signature
The method lacks the this keyword on the first parameter, so it is not an extension method.
Step 2: Verify other syntax elements
Using yield return inside a foreach is valid, and '%' is a valid operator in C#.
Final Answer:
Missing 'this' keyword on the first parameter to make it an extension method. -> Option C
Quick Check:
Extension method requires 'this' on first parameter [OK]
Hint: Extension methods always need 'this' on first parameter [OK]
Common Mistakes:
Forgetting 'this' keyword
Thinking 'yield return' is invalid in loops
Misunderstanding the modulo operator
5. You want to create a custom LINQ extension method ToDictionaryByLength that converts an IEnumerable<string> into a Dictionary<int, List<string>>, grouping strings by their length. Which implementation correctly achieves this?
hard
A. public static Dictionary ToDictionaryByLength(IEnumerable<string> source) {
return source.ToDictionary(s => s.Length, s => new List<string> { s });
}
B. public static Dictionary ToDictionaryByLength(this IEnumerable<string> source) {
return source.GroupBy(s => s.Length)
.ToDictionary(g => g.Key, g => g.ToList());
}
C. public static Dictionary ToDictionaryByLength(this IEnumerable<string> source) {
var dict = new Dictionary();
foreach (var s in source) {
dict[s.Length] = new List<string> { s };
}
return dict;
}
D. public static Dictionary ToDictionaryByLength(this IEnumerable<string> source) {
return source.ToDictionary(s => s.Length, s => s);
}
Solution
Step 1: Understand grouping by length
We need to group strings by their length, so GroupBy with key s.Length is correct.
Step 2: Convert groups to dictionary
Use ToDictionary with group key and list of strings as value.
Step 3: Check other options
public static Dictionary ToDictionaryByLength(IEnumerable<string> source) {
return source.ToDictionary(s => s.Length, s => new List<string> { s });
} does not group; it tries to create dictionary with duplicate keys causing error. public static Dictionary ToDictionaryByLength(this IEnumerable<string> source) {
var dict = new Dictionary();
foreach (var s in source) {
dict[s.Length] = new List<string> { s };
}
return dict;
} overwrites values instead of adding to list. public static Dictionary ToDictionaryByLength(this IEnumerable<string> source) {
return source.ToDictionary(s => s.Length, s => s);
} tries to map to string instead of list.
Final Answer:
public static Dictionary<int, List<string>> ToDictionaryByLength(this IEnumerable<string> source) {
return source.GroupBy(s => s.Length)
.ToDictionary(g => g.Key, g => g.ToList());
} -> Option B
Quick Check:
GroupBy + ToDictionary with list = correct grouping [OK]
Hint: Use GroupBy then ToDictionary with list conversion [OK]
Common Mistakes:
Not grouping before ToDictionary
Overwriting dictionary entries instead of adding to list
Returning wrong value types in dictionary
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