TypeScript Strict Mode and Why It Matters - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output when strict mode is enabled?

Consider this TypeScript code with strict mode enabled. What will be the output when compiled and run?

Typescript

function greet(name: string) {
  return `Hello, ${name.toUpperCase()}`;
}

console.log(greet(undefined as any));

A TypeError at runtime: Cannot read property 'toUpperCase' of undefined

B Hello, UNDEFINED

C Compilation error: Argument of type 'undefined' is not assignable to parameter of type 'string'.

D Hello, undefined

💡 Hint

'any' bypasses type checks, even in strict mode.

The 'as any' cast bypasses TypeScript's type checking, so the code compiles. At runtime, 'name' is undefined, causing a TypeError on toUpperCase().

🧠 Conceptual

intermediate

Why does strictNullChecks matter in TypeScript strict mode?

Which statement best explains the role of the strictNullChecks option in TypeScript strict mode?

A It automatically converts null and undefined to empty strings.

B It allows null and undefined to be assigned to any type without errors.

C It disables all type checking related to null and undefined.

D It requires explicit handling of null and undefined values, preventing accidental runtime errors.

💡 Hint

Think about how strictNullChecks changes how null and undefined are treated in types.

strictNullChecks forces you to explicitly check or handle null and undefined, reducing bugs caused by unexpected null values.

🔧 Debug

advanced

Identify the error caused by strict mode

What error will this TypeScript code produce when strict mode is enabled?

Typescript

let data: { id: number; name?: string } = { id: 1 };

console.log(data.name.length);

A Runtime error: Cannot read property 'length' of undefined.

B Compilation error: Object is possibly 'undefined'.

C No error, outputs undefined.

D Compilation error: Property 'length' does not exist on type 'string | undefined'.

💡 Hint

Check how optional properties are treated in strict mode.

Because 'name' is optional, it might be undefined. Strict mode requires you to check before accessing 'length'.

📝 Syntax

advanced

Which code snippet correctly uses strict mode with non-null assertion?

Given strict mode is enabled, which option correctly accesses a possibly undefined variable without error?

Typescript

let user: { email?: string } = {};
// Access user.email length safely

A console.log(user.email!.length);

B console.log(user.email.length);

C console.log(user.email?.length);

D console.log(user.email.length!);

💡 Hint

Non-null assertion operator tells TypeScript the value is not null or undefined.

Option A uses the non-null assertion operator (!) correctly to tell TypeScript that user.email is not undefined here.

🚀 Application

expert

How many errors will strict mode report in this code?

Count the number of TypeScript compilation errors when strict mode is enabled for this code:

Typescript

function process(input: string | null) {
  console.log(input.toUpperCase());
  let count: number = null;
  let value: any = 5;
  let result: number = value;
}

A 1 error

B 3 errors

C 2 errors

D No errors

💡 Hint

Check each line for type mismatches or unsafe operations under strict mode.

1) input.toUpperCase() is unsafe because input can be null. 2) Assigning null to number is invalid. No error on assigning any to number, as any is assignable to any type. Total: 2 errors.

Practice

1. What is the main purpose of enabling strict mode in TypeScript?

easy

A. To allow JavaScript code to run without any type checking

B. To make TypeScript run faster by skipping type checks

C. To make TypeScript check your code more carefully and catch errors early

D. To disable all compiler warnings and errors

Solution

1. Step 1: Understand what strict mode does

   Strict mode in TypeScript enables more thorough type checking to find mistakes early.

2. Step 2: Compare options with strict mode purpose

   Only To make TypeScript check your code more carefully and catch errors early correctly describes strict mode as making TypeScript check code carefully and catch errors early.

3. Final Answer:

   To make TypeScript check your code more carefully and catch errors early -> Option C

4. Quick Check:

   Strict mode = careful checks [OK]

Hint: Strict mode means more checks, fewer bugs [OK]

Common Mistakes:

• Thinking strict mode disables checks
• Confusing strict mode with performance settings
• Believing strict mode allows ignoring types

2. Which setting in tsconfig.json enables strict mode in TypeScript?

easy

A. "strictCheck": true

B. "strict": true

C. "enableStrict": true

D. "typeCheck": true

Solution

1. Step 1: Identify the correct tsconfig.json key for strict mode

   The official key to enable strict mode is "strict" set to true.

2. Step 2: Eliminate incorrect options

   Options A, B, and C are not valid TypeScript compiler options for strict mode.

3. Final Answer:

   "strict": true -> Option B

4. Quick Check:

   Strict mode key = "strict" [OK]

Hint: Look for "strict": true in tsconfig.json [OK]

Common Mistakes:

• Using incorrect keys like "strictCheck"
• Confusing strict mode with other compiler options
• Forgetting to put the key inside tsconfig.json

3. What will happen if you run this TypeScript code with strict mode enabled?

let name: string;
console.log(name.length);

medium

A. It will cause a compile-time error because 'name' is used before being assigned

B. It will print the length of the string

C. It will print 'undefined'

D. It will run without errors and print 0

Solution

1. Step 1: Understand variable initialization in strict mode

   With strict mode, TypeScript requires variables to be initialized before use.

2. Step 2: Analyze the code behavior

   The variable 'name' is declared but not assigned a value before accessing 'name.length', causing a compile-time error.

3. Final Answer:

   It will cause a compile-time error because 'name' is used before being assigned -> Option A

4. Quick Check:

   Uninitialized variable use = error [OK]

Hint: Uninitialized variables cause errors in strict mode [OK]

Common Mistakes:

• Assuming variables default to empty string
• Thinking code runs without errors
• Confusing runtime errors with compile-time errors

4. Identify the error in this TypeScript code when strict mode is enabled and fix it:

function greet(name: string | undefined) {
  console.log(name.toUpperCase());
}

medium

A. Error: 'name' might be undefined; fix by checking if 'name' exists before calling toUpperCase()

B. Error: 'toUpperCase' is not a function; fix by casting 'name' to string

C. No error, code is correct

D. Error: Missing return type; fix by adding ': void' to function

Solution

1. Step 1: Understand strict null checks

   With strict mode, TypeScript requires checking if a variable that can be undefined is actually defined before using it.

2. Step 2: Identify the error and fix

   Calling 'toUpperCase()' on 'name' without checking if it's undefined causes an error. Fix by adding a check like 'if (name)'.

3. Final Answer:

   Error: 'name' might be undefined; fix by checking if 'name' exists before calling toUpperCase() -> Option A

4. Quick Check:

   Check undefined before method call [OK]

Hint: Check for undefined before using variable methods [OK]

Common Mistakes:

• Ignoring possible undefined values
• Trying to cast without checks
• Assuming no error without strict mode

5. You have this TypeScript code with strict mode enabled:

interface User {
  id: number;
  name?: string;
}

function printUserName(user: User) {
  console.log(user.name.toUpperCase());
}

How can you fix the code to avoid errors and safely print the user's name in uppercase if it exists?

hard

A. Assign a default value: console.log((user.name || '').toUpperCase());

B. Remove the question mark from 'name' in interface to make it required

C. Use a non-null assertion: console.log(user.name!.toUpperCase());

D. Add a check: if (user.name) console.log(user.name.toUpperCase());

Solution

1. Step 1: Understand optional properties and strict mode

   The 'name' property is optional, so it can be undefined. Strict mode requires checking before use.

2. Step 2: Evaluate options to safely access 'name'

   Add a check: if (user.name) console.log(user.name.toUpperCase()); safely checks if 'name' exists before calling 'toUpperCase()', avoiding errors. Options B, C, and D have drawbacks or risks.

3. Final Answer:

   Add a check: if (user.name) console.log(user.name.toUpperCase()); -> Option D

4. Quick Check:

   Check optional before use = safe code [OK]

Hint: Check optional properties before use in strict mode [OK]

Common Mistakes:

• Using non-null assertion without guarantee
• Assuming optional properties are always defined
• Removing optional marker without reason