Optional chaining with types in Typescript - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to safely access the property name of user.

Typescript

const userName = user[1].name;

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

A!

B.

C?.

D&&

Attempts:

3 left

💡 Hint

Common Mistakes

Using dot . directly causes errors if user is undefined.

Using ! asserts non-null but can cause runtime errors if wrong.

Using ?. allows safe access to name even if user is undefined or null.

2fill in blank

medium

Complete the code to safely call the method getAge on person.

Typescript

const age = person[1].getAge();

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

A?.

B&&

C.

D!

Attempts:

3 left

💡 Hint

Common Mistakes

Using ! forces non-null but can cause errors if person is null.

Using dot . directly causes runtime errors if person is undefined.

Using ?. calls getAge only if person is not null or undefined.

3fill in blank

hard

Fix the error in safely accessing city inside address of user.

Typescript

const city = user.address[1].city;

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

A.

B!

C&&

D?.

Attempts:

3 left

💡 Hint

Common Mistakes

Using dot . directly causes runtime errors if address is undefined.

Using ! asserts non-null but can cause errors if wrong.

Use ?. after address to avoid errors if address is undefined.

4fill in blank

hard

Fill both blanks to create a dictionary of user names with their city if available.

Typescript

const userCities = { [user.name]: user.address[1].city [2] 'Unknown' };

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

A?.

B||

C&&

D!

Attempts:

3 left

💡 Hint

Common Mistakes

Using ! incorrectly can cause runtime errors.

Using && instead of || will not provide a default value.

Use ?. to safely access city and || to provide a default value if city is undefined.

5fill in blank

hard

Fill all three blanks to safely access the nested property zipcode or return 'N/A'.

Typescript

const zip = user[1].address[2].zipcode [3] 'N/A';

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

A?.

C||

D!

Attempts:

3 left

💡 Hint

Common Mistakes

Using ! can cause errors if the property is missing.

Using dot . without optional chaining can cause runtime errors.

Use optional chaining twice to safely access nested properties and logical OR to provide a fallback value.

Practice

1. What does the optional chaining operator ?. do in TypeScript?

easy

A. It forces a property to be non-nullable.

B. It throws an error if the property does not exist.

C. It converts a value to a boolean.

D. It safely accesses a property or method, returning undefined if the value is null or undefined.

Solution

1. Step 1: Understand optional chaining purpose

   Optional chaining ?. allows safe access to nested properties or methods without causing runtime errors if the object is null or undefined.

2. Step 2: Identify behavior on null or undefined

   If the value before ?. is null or undefined, it returns undefined instead of throwing an error.

3. Final Answer:

   It safely accesses a property or method, returning undefined if the value is null or undefined. -> Option D

4. Quick Check:

   Optional chaining = safe access without error [OK]

Hint: Optional chaining returns undefined if value is null or undefined [OK]

Common Mistakes:

• Thinking it throws an error on missing property
• Confusing it with non-null assertion operator
• Assuming it converts values to boolean

2. Which of the following is the correct syntax to safely access the property name of an object user using optional chaining?

easy

A. user?.name()

B. user?.name

C. user?.[name]

D. user?.name?

Solution

1. Step 1: Review optional chaining syntax for properties

   The correct syntax to access a property safely is object?.property.

2. Step 2: Check each option

   user?.name uses user?.name which is correct. user?.[name] uses brackets incorrectly. user?.name() treats name as a method call. user?.name? has invalid syntax.

3. Final Answer:

   user?.name -> Option B

4. Quick Check:

   Property access with ?. = user?.name [OK]

Hint: Use ?. before property name without parentheses for safe access [OK]

Common Mistakes:

• Adding parentheses when accessing properties
• Using brackets incorrectly with optional chaining
• Placing ? after property name

3. Consider the following TypeScript code:

type User = { profile?: { age?: number } };
const user: User = {};
const age = user.profile?.age ?? 30;
console.log(age);

What will be the output?

medium

A. 30

B. null

C. Error at runtime

D. undefined

Solution

1. Step 1: Analyze optional chaining and nullish coalescing

   The expression user.profile?.age safely tries to access age. Since profile is undefined, this returns undefined.

2. Step 2: Apply nullish coalescing operator

   The ?? operator returns the right side value 30 because the left side is undefined.

3. Final Answer:

   30 -> Option A

4. Quick Check:

   undefined ?? 30 = 30 [OK]

Hint: ?. returns undefined if missing; ?? provides default value [OK]

Common Mistakes:

• Expecting undefined instead of default value
• Confusing ?? with || operator
• Assuming runtime error on missing property

4. Identify the error in this TypeScript code snippet:

interface Data {
  info?: {
    value: number;
  };
}

const data: Data = {};
const val = data?.info.value;

medium

A. No error, code works fine

B. Runtime error: accessing property of undefined

C. Type error: missing optional chaining before 'value'

D. Syntax error: optional chaining used incorrectly on data

Solution

1. Step 1: Check optional chaining usage

   The code uses data?.info.value. Optional chaining is applied only on data, but info can be undefined.

2. Step 2: Identify missing optional chaining

   Accessing info.value without optional chaining can cause a runtime error if info is undefined. It should be data.info?.value or data?.info?.value.

3. Final Answer:

   Type error: missing optional chaining before 'value' -> Option C

4. Quick Check:

   Optional chaining needed before 'value' to avoid error [OK]

Hint: Chain ?. for each possibly undefined property [OK]

Common Mistakes:

• Applying ?. only once for nested properties
• Assuming optional chaining on parent covers all children
• Ignoring possible undefined intermediate objects

5. Given this nested object type and variable:

type Config = {
  settings?: {
    theme?: {
      color?: string;
    };
  };
};

const config: Config = {};

Which expression safely accesses the color property and provides a default of 'blue' if missing?

hard

A. config?.settings?.theme?.color ?? 'blue'

B. config.settings.theme.color || 'blue'

C. config.settings?.theme.color ?? 'blue'

D. config?.settings.theme?.color || 'blue'

Solution

1. Step 1: Use optional chaining for all nested optional properties

   Since settings, theme, and color are optional, use ?. at each level: config?.settings?.theme?.color.

2. Step 2: Use nullish coalescing for default value

   Use ?? 'blue' to provide default only if the value is null or undefined, avoiding falsey value issues.

3. Final Answer:

   config?.settings?.theme?.color ?? 'blue' -> Option A

4. Quick Check:

   Chain ?. and use ?? for default [OK]

Hint: Chain ?. for each optional and use ?? for default [OK]

Common Mistakes:

• Using || which treats empty string as false
• Missing optional chaining on some nested properties
• Mixing ?. and non-optional access incorrectly