Interface vs type alias decision in Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare an interface named Person with a name property.

Typescript

interface Person [1] {
  name: string;
}

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

A[]

B()

C{}

D<>

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses or angle brackets instead of curly braces.

The correct syntax to declare an interface body is with curly braces {}.

2fill in blank

medium

Complete the code to declare a type alias named Point with x and y as numbers.

Typescript

type Point = [1] {
  x: number;
  y: number;
};

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

A{

B(

Cinterface

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'interface' keyword or parentheses instead of curly braces.

Type aliases use an equals sign followed by an object type enclosed in curly braces {}.

3fill in blank

hard

Fix the error in extending an interface named Animal with a new interface Dog.

Typescript

interface Dog [1] Animal {
  breed: string;
}

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

Aimplements

Buses

Cinherits

Dextends

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'implements' or 'inherits' which are incorrect for interfaces.

Interfaces extend other interfaces using the extends keyword.

4fill in blank

hard

Fill both blanks to create a type alias named ID that can be a string or a number.

Typescript

type ID = [1] | [2];

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

Astring

Bnumber

Cboolean

Dobject

Attempts:

3 left

💡 Hint

Common Mistakes

Using types like boolean or object which are not valid for this ID.

The type alias uses a union type with string or number separated by a pipe |.

5fill in blank

hard

Complete the code to declare an interface named Vehicle with optional wheels and readonly brand properties.

Typescript

interface Vehicle {
   wheels?: number;
  [1] brand: string;
  drive(): void;
}

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

Areadonly

Boptional

Dpublic

Attempts:

3 left

💡 Hint

Common Mistakes

Adding 'optional' keyword which does not exist in TypeScript.

Forgetting the question mark for optional properties.

The wheels property is optional using ? without extra keywords. The brand property is readonly using the readonly keyword. The drive method has no modifier.

Practice

1. Which statement best describes when to use an interface instead of a type alias in TypeScript?

easy

A. Use interface when you want to extend or merge object shapes.

B. Use interface only for primitive types like string or number.

C. Use interface to create union or tuple types.

D. Use interface when you want to name any type, including functions.

Solution

1. Step 1: Understand interface capabilities

   Interfaces are designed to describe object shapes and support extension and merging.

2. Step 2: Compare with type alias features

   Type aliases can name any type but do not support merging and are less suited for extension of object shapes.

3. Final Answer:

   Use interface when you want to extend or merge object shapes. -> Option A

4. Quick Check:

   Interface = extension and merging [OK]

Hint: Interfaces merge and extend; type aliases do not [OK]

Common Mistakes:

• Thinking interfaces can name union types
• Believing type aliases support merging
• Using interface for primitives only

2. Which of the following is the correct syntax to declare a type alias for a union of string literals?

easy

A. interface Status = 'success' | 'error';

B. interface Status: 'success' | 'error';

C. type Status: 'success' | 'error';

D. type Status = 'success' | 'error';

Solution

1. Step 1: Recall type alias syntax

   Type aliases use the keyword type followed by the name, equals sign, and the type expression.

2. Step 2: Identify union type declaration

   The union of string literals is written as 'success' | 'error' after the equals sign.

3. Final Answer:

   type Status = 'success' | 'error'; -> Option D

4. Quick Check:

   Type alias syntax = type Name = TypeExpression [OK]

Hint: Use 'type' keyword with '=' for aliases [OK]

Common Mistakes:

• Using 'interface' for union types
• Using ':' instead of '=' in type alias
• Mixing interface and type alias syntax

3. Given the following code, what will be the output of console.log(user1.age);?

interface User {
  name: string;
}

interface User {
  age: number;
}

const user1: User = { name: 'Alice', age: 30 };

medium

A. 30

B. undefined

C. TypeScript error: Duplicate identifier 'User'

D. Runtime error

Solution

1. Step 1: Understand interface merging

   TypeScript merges multiple interface declarations with the same name into one combined interface.

2. Step 2: Check the combined interface properties

   The merged User interface has both name and age properties, so user1 can have both.

3. Final Answer:

   30 -> Option A

4. Quick Check:

   Interface merging allows combined properties [OK]

Hint: Interfaces with same name merge properties [OK]

Common Mistakes:

• Expecting a duplicate identifier error
• Thinking type aliases merge
• Assuming age is undefined

4. Identify the error in the following code and choose the correct fix:

type Person = {
  name: string;
};

type Person = {
  age: number;
};

medium

A. Use union type instead of redeclaration.

B. Change both to interfaces to allow merging.

C. Type aliases cannot be redeclared; rename one alias.

D. No error; code is valid.

Solution

1. Step 1: Understand redeclaration rules for type aliases

   Type aliases cannot be declared multiple times with the same name; this causes a duplicate identifier error.

2. Step 2: Identify the fix

   Renaming one of the type aliases avoids the conflict; interfaces can merge but type aliases cannot.

3. Final Answer:

   Type aliases cannot be redeclared; rename one alias. -> Option C

4. Quick Check:

   Type aliases = no redeclaration [OK]

Hint: Type aliases cannot be declared twice with same name [OK]

Common Mistakes:

• Assuming type aliases merge like interfaces
• Ignoring duplicate identifier error
• Trying to redeclare type alias without rename

5. You want to create a reusable type that can represent either a user object with name and age or a string ID. Which is the best approach?

hard

A. Use an interface for the user object and a separate type alias for the union with string.

B. Use a type alias with a union of the user object type and string.

C. Use an interface with a union inside it.

D. Use two interfaces and merge them.

Solution

1. Step 1: Identify the need for union types

   You want a type that can be either an object or a string, which requires a union type.

2. Step 2: Choose the best syntax for union types

   Type aliases can represent unions directly, while interfaces cannot represent unions themselves.

3. Final Answer:

   Use a type alias with a union of the user object type and string. -> Option B

4. Quick Check:

   Type aliases = best for unions [OK]

Hint: Use type alias for unions including objects and primitives [OK]

Common Mistakes:

• Trying to use interface for union types
• Merging interfaces to represent unions
• Separating types unnecessarily