Literal types and value narrowing in Typescript - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a variable with a literal type.

Typescript

let direction: 'up' | 'down' = [1];

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

A"left"

B"up"

C"right"

D"forward"

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning a string not in the literal type union.

Forgetting to use quotes around the string.

The variable direction can only be 'up' or 'down'. Assigning 'up' is correct.

2fill in blank

medium

Complete the function to narrow the literal type using a switch statement.

Typescript

function move(direction: 'left' | 'right') {
  switch (direction) {
    case [1]:
      return 'Moving left';
    case 'right':
      return 'Moving right';
  }
}

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

A"left"

B"up"

C"down"

D"forward"

Attempts:

3 left

💡 Hint

Common Mistakes

Using a string not in the union type.

Omitting quotes around the case value.

The switch case must match one of the literal types. 'left' is correct here.

3fill in blank

hard

Fix the error in the type narrowing by completing the if condition.

Typescript

function isYes(answer: 'yes' | 'no'): boolean {
  if (answer === [1]) {
    return true;
  }
  return false;
}

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

A"no"

B"maybe"

C"y"

D"yes"

Attempts:

3 left

💡 Hint

Common Mistakes

Checking against a string not in the union type.

Using loose equality instead of strict equality.

The function returns true only if the answer is exactly 'yes'.

4fill in blank

hard

Fill both blanks to create a function that narrows a literal type and returns a message.

Typescript

function getStatus(status: 'success' | 'error') {
  if (status === [1]) {
    return 'Operation was successful';
  } else if (status === [2]) {
    return 'Operation failed';
  }
}

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

A"success"

B"fail"

C"error"

D"pending"

Attempts:

3 left

💡 Hint

Common Mistakes

Using strings not in the union type.

Mixing up the order of the literals.

The function checks if status is 'success' or 'error' and returns the corresponding message.

5fill in blank

hard

Fill all three blanks to create a function that narrows a literal type and returns a custom message.

Typescript

function respond(action: 'start' | 'stop' | 'pause') {
  if (action === [1]) {
    return 'Starting';
  } else if (action === [2]) {
    return 'Stopping';
  } else if (action === [3]) {
    return 'Pausing';
  }
}

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

A"start"

B"stop"

C"pause"

D"play"

Attempts:

3 left

💡 Hint

Common Mistakes

Using a string not in the union type.

Forgetting quotes around string literals.

The function checks the action literal and returns the matching message.

Practice

1. What does a literal type in TypeScript allow you to do?

easy

A. Specify exact allowed values for a variable

B. Create a new class

C. Define a function return type as any

D. Ignore type checking

Solution

1. Step 1: Understand the purpose of literal types

   Literal types let you restrict a variable to exact values like 'yes' or 10 instead of general types like string or number.

2. Step 2: Compare options with the definition

   Only Specify exact allowed values for a variable correctly describes this feature; others describe unrelated concepts.

3. Final Answer:

   Specify exact allowed values for a variable -> Option A

4. Quick Check:

   Literal types = exact allowed values [OK]

Hint: Literal types fix variable values exactly [OK]

Common Mistakes:

• Confusing literal types with classes
• Thinking literal types disable type checking
• Assuming literal types allow any value

2. Which of the following is the correct way to declare a variable with a literal type of the string 'open' in TypeScript?

easy

A. let status = 'open';

B. let status: string = 'open';

C. let status: 'open' = 'open';

D. let status: 'open'; status = 'closed';

Solution

1. Step 1: Check the syntax for literal type declaration

   To declare a variable with a literal type, you specify the exact value in quotes as the type, like 'open'.

2. Step 2: Evaluate each option

   let status: 'open' = 'open'; correctly declares status with type 'open' and assigns 'open'. let status: string = 'open'; uses string type, not literal. let status = 'open'; infers string type but not literal. let status: 'open'; status = 'closed'; declares type 'open' but assigns 'closed', which is invalid.

3. Final Answer:

   let status: 'open' = 'open'; -> Option C

4. Quick Check:

   Literal type syntax uses exact value in quotes [OK]

Hint: Use quotes around value to declare literal type [OK]

Common Mistakes:

• Assigning a different value than the literal type
• Using general types instead of literal types
• Omitting the type and expecting literal type

3. What will be the output of the following TypeScript code?

type Direction = 'left' | 'right';
function move(dir: Direction) {
  if (dir === 'left') {
    return 'Moving left';
  } else {
    return 'Moving right';
  }
}
console.log(move('right'));

medium

A. Moving right

B. Type error at compile time

C. Moving left

D. Undefined

Solution

1. Step 1: Understand the function behavior with literal type

   The function move accepts only 'left' or 'right'. It returns 'Moving left' if input is 'left', else 'Moving right'.

2. Step 2: Trace the function call with 'right'

   Calling move('right') triggers the else branch, returning 'Moving right'.

3. Final Answer:

   Moving right -> Option A

4. Quick Check:

   Input 'right' returns 'Moving right' [OK]

Hint: Check input matches condition for output [OK]

Common Mistakes:

• Confusing which branch runs for 'right'
• Expecting compile error for valid literal
• Ignoring else branch

4. Identify the error in this TypeScript code using literal types and value narrowing:

type Status = 'success' | 'error';
function handle(status: Status) {
  if (status === 'success') {
    console.log('Success!');
  } else if (status === 'fail') {
    console.log('Failed!');
  }
}

medium

A. Missing return type in function

B. No error, code is correct

C. Variable status is not declared

D. The else if condition checks an invalid literal 'fail'

Solution

1. Step 1: Check the allowed literal types for Status

   Status can only be 'success' or 'error'.

2. Step 2: Analyze the else if condition

   The code checks if status === 'fail', which is not part of Status type, causing a type error.

3. Final Answer:

   The else if condition checks an invalid literal 'fail' -> Option D

4. Quick Check:

   Check conditions must match literal types [OK]

Hint: Check all compared literals are in the type union [OK]

Common Mistakes:

• Using literals not declared in the union type
• Ignoring TypeScript errors on invalid comparisons
• Assuming missing return type causes error here

5. Given the following TypeScript code, what is the type of variable result after the if statement?

type Mode = 'auto' | 'manual';
function setMode(mode: Mode) {
  let result: string | undefined;
  if (mode === 'auto') {
    result = 'Automatic mode set';
  } else {
    result = undefined;
  }
  return result;
}

hard

A. string

B. string | undefined

C. undefined

D. Mode

Solution

1. Step 1: Analyze variable declaration and assignments

   Variable result is declared as string | undefined. Inside if, assigned string; else assigned undefined.

2. Step 2: Determine narrowed type after if

   Since result can be either string or undefined depending on mode, its type remains string | undefined.

3. Final Answer:

   string | undefined -> Option B

4. Quick Check:

   Variable type includes all assigned values [OK]

Hint: Variable type covers all assigned values after narrowing [OK]

Common Mistakes:

• Assuming type narrows to string only
• Ignoring undefined assignment
• Confusing Mode type with result type