Why instanceof fails on interfaces in Typescript - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if 'obj' is an instance of class 'User'.

Typescript

if (obj [1] User) {
  console.log('It is a User');
}

Drag options to blanks, or click blank then click option'

Ainstanceof

Btypeof

Cin

Dequals

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'typeof' to check class instances.

Trying to use 'instanceof' with interfaces.

The instanceof operator checks if an object is an instance of a class or constructor function.

2fill in blank

medium

Complete the code to define an interface 'Person' with a 'name' property.

Typescript

interface Person {
  [1]: string;
}

Drag options to blanks, or click blank then click option'

Aage

Bname

Cid

Demail

Attempts:

3 left

💡 Hint

Common Mistakes

Using properties not defined in the interface.

Confusing property names.

The interface 'Person' requires a 'name' property of type string.

3fill in blank

hard

Fix the error in the code that tries to use 'instanceof' with an interface.

Typescript

function checkPerson(obj: any) {
  if ('name' [1] obj) {
    console.log('Is a Person');
  }
}

Drag options to blanks, or click blank then click option'

Aequals

Btypeof

Cin

Dinstanceof

Attempts:

3 left

💡 Hint

Common Mistakes

Trying to use 'instanceof' with interfaces.

Using 'typeof' to check for interface types.

You cannot use instanceof with interfaces because interfaces do not exist at runtime. Instead, use the in operator to check for property presence.

4fill in blank

hard

Fill the blank to create a type guard function that checks if an object implements the 'Person' interface.

Typescript

function isPerson(obj: any): obj is Person {
  return [1] obj;
}

Drag options to blanks, or click blank then click option'

AhasOwnProperty

Bin

Cobj

D'name' in

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'hasOwnProperty' incorrectly.

Reversing the order of operands in the 'in' operator.

The in operator checks if the property exists in the object or its prototype chain. Here, we check if 'name' is in obj.

5fill in blank

hard

Fill all three blanks to create a safe type guard that checks if 'obj' implements 'Person' by verifying the 'name' property is a string.

Typescript

function isPerson(obj: any): obj is Person {
  return (typeof obj [1] 'object') [2] obj !== null && [3] in obj && typeof obj.name === 'string';
}

Drag options to blanks, or click blank then click option'

A&&

B===

C'name'

D||

Attempts:

3 left

💡 Hint

Common Mistakes

Using logical OR || instead of AND &&.

Checking property presence incorrectly.

Not verifying that 'obj' is not null.

This function first checks that obj is an object and not null, then checks if it has a 'name' property, and finally verifies that 'name' is a string.

Practice

1. Why does the instanceof operator fail when used to check if an object implements a TypeScript interface?

easy

A. Because interfaces are only used for styling in TypeScript.

B. Because instanceof only works with primitive types.

C. Because interfaces do not exist at runtime in JavaScript.

D. Because instanceof requires a function as the right operand.

Solution

1. Step 1: Understand what instanceof checks

   The instanceof operator checks if an object was created from a specific class or constructor function at runtime.

2. Step 2: Recognize interface behavior in TypeScript

   Interfaces in TypeScript are only used during compilation and do not exist in the generated JavaScript code, so instanceof cannot check them.

3. Final Answer:

   Because interfaces do not exist at runtime in JavaScript. -> Option C

4. Quick Check:

   Interfaces disappear at runtime = A [OK]

Hint: Remember: interfaces vanish after compile, so instanceof fails [OK]

Common Mistakes:

• Thinking instanceof works with interfaces
• Confusing interfaces with classes
• Assuming interfaces exist at runtime
• Using instanceof with primitives

2. Which of the following is the correct syntax to check if an object is an instance of a class Person in TypeScript?

easy

A. if (obj instanceof Person) { }

B. if (obj instanceof 'Person') { }

C. if (obj instanceof InterfacePerson) { }

D. if (obj instanceof { name: string }) { }

Solution

1. Step 1: Recall instanceof syntax

   The right side of instanceof must be a class or constructor function, not a string or interface.

2. Step 2: Identify valid usage

   Only obj instanceof Person is valid syntax to check if obj is created from class Person.

3. Final Answer:

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

4. Quick Check:

   Right operand must be class/constructor = C [OK]

Hint: Use class name without quotes for instanceof [OK]

Common Mistakes:

• Using interface names with instanceof
• Putting class name in quotes
• Using object literals on right side
• Confusing type and value syntax

3. Consider this TypeScript code:

interface Animal { name: string }
class Dog implements Animal { constructor(public name: string) {} }
const pet = new Dog('Buddy');
console.log(pet instanceof Dog);
console.log(pet instanceof Animal);

What is the output?

medium

A. false\nfalse

B. true\ntrue

C. false\ntrue

D. true\nfalse

Solution

1. Step 1: Check pet instanceof Dog

   Since pet was created with new Dog(), this returns true.

2. Step 2: Check pet instanceof Animal

   Animal is an interface and does not exist at runtime, so this returns false.

3. Final Answer:

   true\nfalse -> Option D

4. Quick Check:

   Class instance true, interface instance false = A [OK]

Hint: Only classes work with instanceof, interfaces do not [OK]

Common Mistakes:

• Expecting interface check to return true
• Confusing interface with class
• Assuming interfaces exist at runtime
• Misreading output order

4. Given this code snippet:

interface Vehicle { wheels: number }
function isVehicle(obj: any) {
  return obj instanceof Vehicle;
}

Why does this code cause an error or unexpected behavior?

medium

A. Because Vehicle is an interface and cannot be used with instanceof.

B. Because obj is not typed correctly.

C. Because instanceof requires a string argument.

D. Because Vehicle is not imported.

Solution

1. Step 1: Identify the use of instanceof with interface

   The code uses instanceof with Vehicle, which is an interface.

2. Step 2: Understand interface runtime behavior

   Interfaces do not exist at runtime, so instanceof Vehicle is invalid and causes an error or always returns false.

3. Final Answer:

   Because Vehicle is an interface and cannot be used with instanceof. -> Option A

4. Quick Check:

   Interfaces can't be checked with instanceof = D [OK]

Hint: Never use instanceof with interfaces, only classes [OK]

Common Mistakes:

• Trying to check interfaces with instanceof
• Ignoring runtime absence of interfaces
• Assuming instanceof works with types
• Misunderstanding error cause

5. You want to check at runtime if an object matches the shape of an interface Person { name: string; age: number; }. Since instanceof does not work with interfaces, which approach is best?

hard

A. Use instanceof Person anyway; it will work after compilation.

B. Manually check if object has required properties with correct types.

C. Convert the interface to a class and use instanceof.

D. Use typeof obj === 'Person' to check the type.

Solution

1. Step 1: Understand interface runtime absence

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

2. Step 2: Choose property checks for runtime validation

   The best way is to manually check if the object has the required properties with the expected types.

3. Final Answer:

   Manually check if object has required properties with correct types. -> Option B

4. Quick Check:

   Manual property checks replace instanceof for interfaces = B [OK]

Hint: Check properties manually to mimic interface checks at runtime [OK]

Common Mistakes:

• Using instanceof with interfaces
• Assuming typeof works for interfaces
• Converting interfaces to classes unnecessarily
• Ignoring runtime type checks