Literal types and value narrowing in Typescript

Introduction

Literal types let you use exact values as types. Value narrowing helps TypeScript understand what exact value a variable holds, so it can catch mistakes early.

When you want a variable to only hold a few specific values, like 'yes' or 'no'.

When you want to check a variable's value and have TypeScript know exactly which case it is.

When you want safer code by preventing unexpected values.

When writing functions that behave differently based on exact input values.

Syntax

Typescript

let direction: 'left' | 'right' | 'up' | 'down';

function move(dir: 'left' | 'right' | 'up' | 'down') {
  if (dir === 'left') {
    // do something
  } else if (dir === 'right') {
    // do something else
  }
}

Literal types are written as exact values in quotes, like 'left' or 42.

Value narrowing happens when you check a variable's value with if or switch, so TypeScript knows the exact value inside that block.

Examples

This variable can only be 'yes' or 'no'. Assigning 'maybe' causes an error.

Typescript

let answer: 'yes' | 'no';
answer = 'yes'; // OK
answer = 'maybe'; // Error: not allowed

Inside the if, TypeScript knows dir is exactly 'left'. In else, it knows dir is 'right'.

Typescript

function printDirection(dir: 'left' | 'right') {
  if (dir === 'left') {
    console.log('Going left');
  } else {
    console.log('Going right');
  }
}

Using as const makes the variable a literal type. The switch narrows the value to each case.

Typescript

const status = 'success' as const;

function handleStatus(s: 'success' | 'error') {
  switch (s) {
    case 'success':
      console.log('It worked!');
      break;
    case 'error':
      console.log('There was an error.');
      break;
  }
}

handleStatus(status);

Sample Program

This program uses a literal type Direction to allow only four directions. The function move checks the exact value and prints a message accordingly.

Typescript

type Direction = 'north' | 'south' | 'east' | 'west';

function move(direction: Direction) {
  if (direction === 'north') {
    console.log('Moving up');
  } else if (direction === 'south') {
    console.log('Moving down');
  } else {
    console.log(`Moving ${direction}`);
  }
}

move('north');
move('west');

Important Notes

Literal types help catch errors early by limiting allowed values.

Value narrowing works well with if, else, and switch statements.

Use as const to make variables have literal types instead of general types.

Summary

Literal types let you specify exact allowed values for variables.

Value narrowing helps TypeScript understand which exact value a variable has after checks.

Together, they make your code safer and easier to understand.

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