Type patterns in C Sharp (C#)

Introduction

Type patterns help you check if a value is a certain type and use it right away. This makes your code simpler and clearer.

When you want to check if an object is a specific type before using it.

When you want to safely convert an object to a type and use it in one step.

When you want to write cleaner code instead of separate type checks and casts.

When handling different types in a switch statement or if-else blocks.

When you want to avoid errors from wrong type conversions.

Syntax

C Sharp (C#)

if (variable is TypeName variableName) {
    // use variableName here
}

The is keyword checks the type and creates a new variable if it matches.

This new variable is only available inside the if-block.

Examples

Check if obj is a string and use it as s inside the block.

C Sharp (C#)

object obj = "hello";
if (obj is string s) {
    Console.WriteLine(s.ToUpper());
}

Check if obj is an int and add 10 to it.

C Sharp (C#)

object obj = 123;
if (obj is int number) {
    Console.WriteLine(number + 10);
}

Use type patterns in a switch to handle different types.

C Sharp (C#)

object obj = 3.14;
switch (obj) {
    case double d:
        Console.WriteLine($"Double: {d}");
        break;
    case int i:
        Console.WriteLine($"Int: {i}");
        break;
    default:
        Console.WriteLine("Unknown type");
        break;
}

Sample Program

This program checks each item in an array and prints its type and value using type patterns.

C Sharp (C#)

using System;

class Program {
    static void Main() {
        object[] items = { "apple", 42, 3.14, null };

        foreach (var item in items) {
            if (item is string text) {
                Console.WriteLine($"String: {text}");
            } else if (item is int number) {
                Console.WriteLine($"Integer: {number}");
            } else if (item is double dbl) {
                Console.WriteLine($"Double: {dbl}");
            } else {
                Console.WriteLine("Unknown or null");
            }
        }
    }
}

Important Notes

Type patterns combine type checking and casting in one step.

They help avoid errors from wrong casts and make code easier to read.

Summary

Type patterns check and use a value's type in one step.

They simplify code by removing separate casts.

Useful in if statements and switch cases.

Practice

1. What does a type pattern in C# do when used in an if statement like if (obj is string s)?

easy

A. Always returns true regardless of obj type

B. Converts obj to a string without checking

C. Checks if obj is a string and assigns it to s if true

D. Throws an exception if obj is not a string

Solution

1. Step 1: Understand the is keyword with type pattern

   The is keyword checks if the object is of the specified type.

2. Step 2: Recognize the pattern variable assignment

   If the check is true, the object is assigned to the variable s of that type.

3. Final Answer:

   Checks if obj is a string and assigns it to s if true -> Option C

4. Quick Check:

   Type pattern = Check and assign [OK]

Hint: Type pattern checks type and assigns in one step [OK]

Common Mistakes:

• Thinking it converts without checking
• Assuming it always returns true
• Expecting an exception on type mismatch

2. Which of the following is the correct syntax for using a type pattern in an if statement in C#?

easy

A. if (obj as int number) { }

B. if (int obj is number) { }

C. if (obj int is number) { }

D. if (obj is int number) { }

Solution

1. Step 1: Recall correct type pattern syntax

   The correct syntax is if (variable is Type name).

2. Step 2: Check each option

   if (obj is int number) { } matches the correct syntax. Options B, C, and D have incorrect order or keywords.

3. Final Answer:

   if (obj is int number) { } -> Option D

4. Quick Check:

   Correct syntax = if (obj is Type name) [OK]

Hint: Remember: if (obj is Type var) [OK]

Common Mistakes:

• Swapping type and variable order
• Using 'as' instead of 'is' for pattern
• Placing keywords incorrectly

3. What will be the output of this C# code?

object obj = 42;
if (obj is int number)
{
    Console.WriteLine(number + 10);
}
else
{
    Console.WriteLine("Not an int");
}

medium

A. 52

B. Not an int

C. Compilation error

D. Runtime exception

Solution

1. Step 1: Check the type of obj

   obj holds the integer 42, so obj is int is true.

2. Step 2: Calculate the output inside the if block

   The variable number is assigned 42, so number + 10 equals 52, which is printed.

3. Final Answer:

   52 -> Option A

4. Quick Check:

   Type check true, output = 42 + 10 = 52 [OK]

Hint: If type matches, variable holds value for use [OK]

Common Mistakes:

• Expecting 'Not an int' output
• Confusing type check with cast
• Thinking code throws error

4. Identify the error in this code snippet using type patterns:

object obj = "hello";
if (obj is int number)
{
    Console.WriteLine(number);
}

medium

A. No output because obj is not an int

B. Compilation error due to missing else block

C. No error, code runs and prints 'hello'

D. Runtime exception when casting string to int

Solution

1. Step 1: Analyze the type check condition

   obj is a string, so obj is int number is false.

2. Step 2: Understand the effect on output

   Since the if condition is false and no else block exists, nothing is printed.

3. Final Answer:

   No output because obj is not an int -> Option A

4. Quick Check:

   False type check = no output [OK]

Hint: False type check skips block, no output [OK]

Common Mistakes:

• Expecting runtime exception
• Thinking else block is mandatory
• Assuming it prints the string

5. Given this method using type patterns, what will it return?
<pre>string Describe(object input)
{
return input switch
{
int i => $"Integer: {i}",
string s => $"String: {s}",
null => "Null value",
_ => "Unknown type"
};
}

What is the result of Describe(3.14)?

hard

A. "Integer: 3"

B. "Unknown type"

C. "String: 3.14"

D. Compilation error

Solution

1. Step 1: Check the switch cases for input 3.14

   3.14 is a double, which does not match int, string, or null.

2. Step 2: Identify the default case

   The underscore _ matches any other type, so it returns "Unknown type".

3. Final Answer:

   "Unknown type" -> Option B

4. Quick Check:

   Switch default case = "Unknown type" [OK]

Hint: Switch default (_) handles unmatched types [OK]

Common Mistakes:

• Assuming 3.14 is int or string
• Expecting compilation error
• Ignoring default case