instanceof type guards in Typescript - Mini Project: Build & Apply

Using instanceof Type Guards in TypeScript

📖 Scenario: You are building a simple program that handles different types of animals. Each animal can make a sound, but the way to get the sound depends on the animal type. You want to safely check the type of each animal before calling its specific method.

🎯 Goal: Create a TypeScript program that uses instanceof type guards to identify animal types and call their specific sound methods.

📋 What You'll Learn

Create two classes: Dog and Cat with a method each to make a sound

Create an array called animals containing instances of Dog and Cat

Use a for loop with instanceof to check each animal's type

Print the sound each animal makes using the correct method

💡 Why This Matters

🌍 Real World

Type guards like <code>instanceof</code> help programs safely handle different object types, common in apps that work with many data models.

💼 Career

Understanding type guards is important for writing safe and clear TypeScript code, a skill valued in frontend and backend development jobs.

1

Create Dog and Cat classes

Create a class called Dog with a method bark() that returns the string "Woof!". Also create a class called Cat with a method meow() that returns the string "Meow!".

Typescript

# Your code here

Hint

Use class keyword to create classes. Define methods inside classes that return the correct strings.

2

Create animals array with Dog and Cat instances

Create an array called animals that contains one instance of Dog and one instance of Cat.

Typescript

class Dog {
    bark(): string {
        return "Woof!";
    }
}

class Cat {
    meow(): string {
        return "Meow!";
    }
}

// Your code here

Hint

Create the array with const animals = [new Dog(), new Cat()].

3

Use instanceof to check animal types

Write a for loop that iterates over animals using the variable animal. Inside the loop, use instanceof Dog to check if animal is a Dog and call animal.bark(). Use instanceof Cat to check if animal is a Cat and call animal.meow().

Typescript

class Dog {
    bark(): string {
        return "Woof!";
    }
}

class Cat {
    meow(): string {
        return "Meow!";
    }
}

const animals = [new Dog(), new Cat()];

// Your code here

Hint

Use for (const animal of animals) and inside use if (animal instanceof Dog) and else if (animal instanceof Cat).

4

Print the sounds made by each animal

Run the program so it prints the sounds made by each animal in the animals array using the console.log statements inside the for loop.

Typescript

class Dog {
    bark(): string {
        return "Woof!";
    }
}

class Cat {
    meow(): string {
        return "Meow!";
    }
}

const animals = [new Dog(), new Cat()];

for (const animal of animals) {
    if (animal instanceof Dog) {
        console.log(animal.bark());
    } else if (animal instanceof Cat) {
        console.log(animal.meow());
    }
}

// Your code here

Hint

Run the program and check the console output shows Woof! and Meow! on separate lines.

Practice

1. What does the instanceof operator check in TypeScript?

easy

A. If a variable is a primitive type

B. If an object is an instance of a specific class

C. If a function returns a boolean

D. If a type is an interface

Solution

1. Step 1: Understand the purpose of instanceof

   The instanceof operator checks whether an object was created from a particular class.

2. Step 2: Differentiate from other checks

   It does not check primitive types, function returns, or interfaces, only class instances.

3. Final Answer:

   If an object is an instance of a specific class -> Option B

4. Quick Check:

   instanceof checks class instance = A [OK]

Hint: Remember: instanceof works only with classes, not interfaces [OK]

Common Mistakes:

• Thinking instanceof works with interfaces
• Confusing instanceof with typeof
• Assuming instanceof checks primitive types

2. Which of the following is the correct syntax to check if obj is an instance of class Car?

easy

A. if (obj instanceof Car) { }

B. if (obj instance Car) { }

C. if (obj is Car) { }

D. if (obj type Car) { }

Solution

1. Step 1: Recall correct instanceof syntax

   The correct syntax uses the keyword instanceof between the object and the class name.

2. Step 2: Identify invalid options

   Options B, C, and D use incorrect keywords or missing syntax.

3. Final Answer:

   if (obj instanceof Car) { } -> Option A

4. Quick Check:

   Correct syntax uses instanceof keyword = A [OK]

Hint: Use 'instanceof' keyword exactly between object and class [OK]

Common Mistakes:

• Using 'is' or 'type' instead of 'instanceof'
• Omitting parentheses around condition
• Writing 'instance' instead of 'instanceof'

3. What will be the output of this code?

class Dog { bark() { return 'woof'; } }
class Cat { meow() { return 'meow'; } }
function speak(animal: Dog | Cat) {
  if (animal instanceof Dog) {
    return animal.bark();
  } else {
    return animal.meow();
  }
}
console.log(speak(new Cat()));

medium

A. 'woof'

B. TypeError at runtime

C. 'meow'

D. undefined

Solution

1. Step 1: Identify the object passed

   The function speak is called with new Cat(), so animal is a Cat instance.

2. Step 2: Check the instanceof condition

   The condition animal instanceof Dog is false, so the else block runs, returning animal.meow().

3. Final Answer:

   'meow' -> Option C

4. Quick Check:

   Cat instance triggers else block = D [OK]

Hint: Trace instanceof condition to pick correct method call [OK]

Common Mistakes:

• Assuming instanceof works with interfaces
• Confusing which method runs for which class
• Expecting 'woof' when passing Cat instance

4. Identify the error in this code snippet:

interface Bird { fly(): void; }
class Eagle implements Bird { fly() { console.log('Flying'); } }
function check(bird: Bird) {
  if (bird instanceof Eagle) {
    bird.fly();
  }
}

medium

A. No error, code is correct

B. Missing return type in fly() method

C. Cannot use instanceof with interfaces

D. Eagle class must be abstract

Solution

1. Step 1: Understand instanceof usage

   The instanceof operator is used with the class Eagle, which is valid even though bird is typed as the interface Bird.

2. Step 2: Verify no errors

   There are no syntax errors, type mismatches, or runtime issues. TypeScript compiles this code and narrows bird to Eagle inside the if block.

3. Final Answer:

   No error, code is correct -> Option A

4. Quick Check:

   instanceof with class on interface-typed param works = D [OK]

Hint: Remember: instanceof works with classes even on interface-typed variables [OK]

Common Mistakes:

• Thinking instanceof can't be used on interface-typed variables
• Confusing instanceof with direct interface checks
• Thinking implementing class must be abstract

5. Given these classes:

class Animal { name: string; constructor(name: string) { this.name = name; } }
class Fish extends Animal { swim() { return 'swimming'; } }
class Bird extends Animal { fly() { return 'flying'; } }

Write a function move that takes animal: Animal and returns 'swimming' if it's a Fish, 'flying' if it's a Bird, or 'unknown' otherwise. Which implementation is correct?

hard

A. function move(animal: Animal) { if (animal instanceof Animal) return 'unknown'; else if (animal instanceof Fish) return animal.swim(); else return animal.fly(); }

B. function move(animal: Animal) { if (animal instanceof Fish) return 'flying'; else if (animal instanceof Bird) return 'swimming'; else return 'unknown'; }

C. function move(animal: Animal) { if (animal instanceof Fish) return animal.fly(); else if (animal instanceof Bird) return animal.swim(); else return 'unknown'; }

D. function move(animal: Animal) { if (animal instanceof Fish) return animal.swim(); else if (animal instanceof Bird) return animal.fly(); else return 'unknown'; }

Solution

1. Step 1: Check instanceof order and method calls

   The function must check if animal is a Fish or Bird using instanceof and call the correct method.

2. Step 2: Validate each option

   function move(animal: Animal) { if (animal instanceof Fish) return animal.swim(); else if (animal instanceof Bird) return animal.fly(); else return 'unknown'; } correctly checks Fish first, calls swim(), then Bird with fly(), else returns 'unknown'. function move(animal: Animal) { if (animal instanceof Animal) return 'unknown'; else if (animal instanceof Fish) return animal.swim(); else return animal.fly(); } wrongly checks Animal first, which always matches. function move(animal: Animal) { if (animal instanceof Fish) return animal.fly(); else if (animal instanceof Bird) return animal.swim(); else return 'unknown'; } swaps methods incorrectly. function move(animal: Animal) { if (animal instanceof Fish) return 'flying'; else if (animal instanceof Bird) return 'swimming'; else return 'unknown'; } returns incorrect strings directly without calling methods.

3. Final Answer:

   function move(animal: Animal) { if (animal instanceof Fish) return animal.swim(); else if (animal instanceof Bird) return animal.fly(); else return 'unknown'; } -> Option D

4. Quick Check:

   Correct instanceof checks and method calls = B [OK]

Hint: Check instanceof order and call matching methods [OK]

Common Mistakes:

• Checking base class first, blocking subclass checks
• Calling wrong methods for classes
• Returning strings instead of calling methods