Why instanceof fails on interfaces in Typescript - See It in Action

Why instanceof Fails on Interfaces

📖 Scenario: Imagine you are building a simple app where you want to check if an object follows a certain shape (interface) before using it.

🎯 Goal: You will learn why the instanceof operator does not work with interfaces in TypeScript and how to check object types safely.

📋 What You'll Learn

Create an interface called Person with properties name and age

Create a class called Employee that implements Person

Create an object of type Employee

Try to use instanceof Person and see why it fails

Use a type guard function to check if an object matches the Person interface

💡 Why This Matters

🌍 Real World

When working with TypeScript, you often want to check if objects have certain shapes before using them safely.

💼 Career

Understanding how to check types correctly helps prevent bugs and is important for writing reliable TypeScript code in professional projects.

1

Create the Person interface and Employee class

Create an interface called Person with properties name (string) and age (number). Then create a class called Employee that implements Person and has a constructor to set name and age.

Typescript

// Your code here

Need a hint?

Interfaces define a shape but do not exist at runtime. Classes create real objects you can check with instanceof.

2

Create an Employee object

Create a variable called emp and assign it a new Employee object with name as "Alice" and age as 30.

Typescript

interface Person {
    name: string;
    age: number;
}

class Employee implements Person {
    constructor(public name: string, public age: number) {}
}

// Your code here

Need a hint?

Use the new keyword to create an instance of the class.

3

Try using instanceof with the interface

Write an if statement that checks if emp instanceof Person and inside it, write console.log("emp is a Person"). Also add an else block with console.log("emp is NOT a Person").

Typescript

interface Person {
    name: string;
    age: number;
}

class Employee implements Person {
    constructor(public name: string, public age: number) {}
}

const emp = new Employee("Alice", 30);

// Your code here

Need a hint?

Interfaces do not exist at runtime, so instanceof cannot check them.

4

Use a type guard function to check the interface

Create a function called isPerson that takes obj: any and returns obj is Person. Inside, check if obj has name and age properties with correct types. Then use if (isPerson(emp)) to print "emp is a Person", else print "emp is NOT a Person".

Typescript

interface Person {
    name: string;
    age: number;
}

class Employee implements Person {
    constructor(public name: string, public age: number) {}
}

const emp = new Employee("Alice", 30);

if (emp instanceof Person) {
    console.log("emp is a Person");
} else {
    console.log("emp is NOT a Person");
}

// Your code here

Need a hint?

Type guard functions check properties manually because interfaces vanish after compiling.