Why generics are needed in Typescript - See It in Action

Why generics are needed

📖 Scenario: Imagine you are building a simple storage box system. You want to store different types of items safely without mixing them up or losing their type information.

🎯 Goal: You will create a generic storage box function that can hold any type of item and return it safely, showing why generics help keep code flexible and type-safe.

📋 What You'll Learn

Create a function called storeItem that takes one parameter called item and returns it.

Create a variable called numberBox that stores a number using storeItem.

Create a variable called stringBox that stores a string using storeItem.

Use a generic type parameter T in the storeItem function to keep the type of item.

Print the values of numberBox and stringBox.

💡 Why This Matters

🌍 Real World

Generics are used in many programming tasks where you want to write reusable code that works with different data types, like storing user data, handling lists, or working with APIs.

💼 Career

Understanding generics is important for writing safe and reusable code in TypeScript, which is widely used in web development jobs.

1

Create a simple function to store an item

Create a function called storeItem that takes one parameter called item of type any and returns it.

Typescript

// Create a function called storeItem with parameter item of type any
// Your code here

Hint

Use function storeItem(item: any) { return item; } to accept any type.

2

Store a number and a string using the function

Create a variable called numberBox and assign it the result of calling storeItem with the number 42. Then create a variable called stringBox and assign it the result of calling storeItem with the string 'hello'.

Typescript

function storeItem(item: any) {
    return item;
}

// Create numberBox with storeItem(42)
// Create stringBox with storeItem('hello')
// Your code here

Hint

Use const numberBox = storeItem(42); and const stringBox = storeItem('hello');.

3

Make the function generic to keep the item type

Change the storeItem function to use a generic type parameter T. The parameter item should be of type T and the function should return type T.

Typescript

function storeItem(item: any) {
    return item;
}

const numberBox = storeItem(42);
const stringBox = storeItem('hello');

// Change storeItem to use generic type T
// Your code here

Hint

Use function storeItem<T>(item: T): T { return item; } to keep the type.

4

Print the stored values to see the result

Write two console.log statements to print numberBox and stringBox.

Typescript

function storeItem(item: T): T {
    return item;
}

const numberBox = storeItem(42);
const stringBox = storeItem('hello');

// Print numberBox
// Print stringBox
// Your code here

Hint

Use console.log(numberBox); and console.log(stringBox); to show the values.

Practice

1. Why do we use generics in TypeScript?
function identity<T>(arg: T): T { return arg; }

easy

A. To make code run faster by skipping type checks

B. To force all variables to be of type string

C. To write functions that work with any data type while keeping type safety

D. To avoid using functions altogether

Solution

1. Step 1: Understand the generic function syntax

   The function identity uses <T> to accept any type and return the same type.

2. Step 2: Recognize the benefit of generics

   This allows the function to work with any type while keeping type safety, preventing errors.

3. Final Answer:

   To write functions that work with any data type while keeping type safety -> Option C

4. Quick Check:

   Generics = reusable + type safe functions [OK]

Hint: Generics = flexible + safe code for many types [OK]

Common Mistakes:

• Thinking generics make code faster
• Believing generics force string types
• Confusing generics with avoiding functions

2. Which of the following is the correct syntax to declare a generic function in TypeScript?

easy

A. function example<T>(arg: T): T { return arg; }

B. function example(arg: T): T { return arg; }

C. function example<T>(arg): T { return arg; }

D. function example<T>(arg: T) { return arg;

Solution

1. Step 1: Check generic declaration and usage

   function example<T>(arg: T): T { return arg; } correctly declares <T> and uses arg: T with return type T.

2. Step 2: Identify syntax errors in other options

   function example(arg: T): T { return arg; } misses generic declaration. function example<T>(arg): T { return arg; } misses type for arg. function example<T>(arg: T) { return arg; } misses return type.

3. Final Answer:

   function example<T>(arg: T): T { return arg; } -> Option A

4. Quick Check:

   Generic syntax requires <T> and typed parameters [OK]

Hint: Generic syntax: function name<T>(arg: T): T [OK]

Common Mistakes:

• Omitting <T> after function name
• Not typing function parameters
• Forgetting return type annotation

3. What will be the output of this TypeScript code?

function wrapInArray<T>(value: T): T[] { return [value]; }
const result = wrapInArray(5);
console.log(result);

medium

A. 5

B. [5]

C. Error: Type mismatch

D. ["5"]

Solution

1. Step 1: Understand the generic function behavior

   The function takes a value of any type and returns an array containing that value.

2. Step 2: Apply the function to the number 5

   Calling wrapInArray(5) returns an array with one element: 5.

3. Final Answer:

   [5] -> Option B

4. Quick Check:

   wrapInArray(5) = [5] [OK]

Hint: Generic function returns array with input value [OK]

Common Mistakes:

• Expecting single value instead of array
• Confusing number 5 with string "5"
• Thinking it causes a type error

4. Identify the error in this generic function and fix it:

function getFirstElement<T>(arr: T) {
  return arr[0];
}

medium

A. Parameter should be an array type: arr: T[]

B. Return type should be T[]

C. Function should not use generics here

D. Use arr.get(0) instead of arr[0]

Solution

1. Step 1: Analyze parameter type

   The parameter arr is typed as T, but it should be an array of T to use arr[0].

2. Step 2: Correct the parameter type

   Change arr: T to arr: T[] so the function accepts an array.

3. Final Answer:

   Parameter should be an array type: arr: T[] -> Option A

4. Quick Check:

   Array parameter needs T[] type [OK]

Hint: Array parameters need T[] type, not T [OK]

Common Mistakes:

• Using T instead of T[] for arrays
• Changing return type incorrectly
• Trying to use non-array methods on arrays

5. You want to create a generic function that accepts an array of any type and returns an object mapping each element to its index. Which is the correct implementation?

hard

A. function mapToIndex<T>(arr: T[]): T[] {
return arr.map((item, index) => index);
}

B. function mapToIndex<T>(arr: T): { [key: string]: number } {
const result = {};
arr.forEach((item, index) => { result[item] = index; });
return result;
}

C. function mapToIndex(arr: any[]): any {
const result = {};
arr.forEach((item, index) => { result[item] = index; });
return result;
}

D. function mapToIndex<T>(arr: T[]): { [key: string]: number } {
const result: { [key: string]: number } = {};
arr.forEach((item, index) => { result[String(item)] = index; });
return result;
}

Solution

1. Step 1: Check parameter and return types

   The function should accept an array of type T[] and return an object mapping strings to numbers.

2. Step 2: Verify implementation details

   function mapToIndex<T>(arr: T[]): { [key: string]: number } { const result: { [key: string]: number } = {}; arr.forEach((item, index) => { result[String(item)] = index; }); return result; } converts each item to string as key and assigns its index as value, correctly building the object.

3. Step 3: Identify errors in other options

   function mapToIndex<T>(arr: T): { [key: string]: number } { const result = {}; arr.forEach((item, index) => { result[item] = index; }); return result; } uses arr: T which is incorrect for arrays. function mapToIndex(arr: any[]): any { const result = {}; arr.forEach((item, index) => { result[item] = index; }); return result; } uses any losing type safety. function mapToIndex<T>(arr: T[]): T[] { return arr.map((item, index) => index); } returns an array of indexes, not the mapping object.

4. Final Answer:

   function mapToIndex<T>(arr: T[]): { [key: string]: number } { const result: { [key: string]: number } = {}; arr.forEach((item, index) => { result[String(item)] = index; }); return result; } -> Option D

5. Quick Check:

   Generic array input + object output with string keys [OK]

Hint: Use T[] for arrays and map keys as strings [OK]

Common Mistakes:

• Using T instead of T[] for array parameter
• Returning wrong type (array instead of object)
• Not converting keys to string