Why Union type syntax and behavior in Typescript? - Purpose & Use Cases

The Big Idea

What if you could tell your program to accept different kinds of data without messy checks everywhere?

The Scenario

Imagine you are writing a program that accepts user input which could be either a number or a string. Without union types, you have to write separate code blocks to handle each type manually, checking and converting types everywhere.

The Problem

This manual checking is slow and error-prone. You might forget a check, causing bugs or crashes. The code becomes long and hard to read, making it difficult to maintain or update.

The Solution

Union types let you declare a variable that can hold multiple types in a clean way. The TypeScript compiler helps you by checking your code and guiding you to handle each type properly, reducing errors and making your code simpler.

Before vs After

✗ Before

function process(input: any) {
  if (typeof input === 'string') {
    // handle string
  } else if (typeof input === 'number') {
    // handle number
  }
}

✓ After

function process(input: string | number) {
  if (typeof input === 'string') {
    // handle string
  } else {
    // handle number
  }
}

What It Enables

It enables writing safer and clearer code that naturally handles multiple types without extra boilerplate.

Real Life Example

For example, a function that accepts either a user ID as a number or a username as a string can be typed with a union, making the code easier to understand and less buggy.

Key Takeaways

Manual type checks are slow and error-prone.

Union types let variables hold multiple types cleanly.

They help write safer, clearer, and easier-to-maintain code.

Practice

1. What does a union type in TypeScript allow a variable to do?

easy

A. Hold values of multiple specified types safely

B. Hold only one fixed type value

C. Automatically convert types at runtime

D. Restrict values to a single primitive type

Solution

1. Step 1: Understand union type purpose

   A union type lets a variable accept values from multiple specified types, not just one.

2. Step 2: Compare options with definition

   Only Hold values of multiple specified types safely correctly describes this behavior; others describe fixed or automatic conversions which union types do not do.

3. Final Answer:

   Hold values of multiple specified types safely -> Option A

4. Quick Check:

   Union type = multiple types allowed [OK]

Hint: Union types use | to allow multiple types [OK]

Common Mistakes:

• Thinking union types convert values automatically
• Believing union types restrict to one type only
• Confusing union with intersection types

2. Which of the following is the correct syntax to declare a variable value that can be either a string or a number?

easy

A. let value: string | number;

B. let value: string & number;

C. let value: string, number;

D. let value: string | number[];

Solution

1. Step 1: Recall union type syntax

   Union types use the pipe symbol | between types without parentheses needed.

2. Step 2: Evaluate each option

   let value: string | number; uses correct syntax: string | number. let value: string & number; uses & which means intersection, not union. let value: string, number; uses comma which is invalid. let value: string | number[]; uses number[] which is an array of numbers, not a single number.

3. Final Answer:

   let value: string | number; -> Option A

4. Quick Check:

   Union syntax uses | without commas [OK]

Hint: Use | between types, no commas or & for union [OK]

Common Mistakes:

• Using & instead of | for union types
• Separating types with commas
• Using number[] instead of number

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

function printId(id: string | number) {
  if (typeof id === 'string') {
    console.log('ID is a string: ' + id.toUpperCase());
  } else {
    console.log('ID is a number: ' + id.toFixed(2));
  }
}
printId(42);

medium

A. ID is a string: 42

B. ID is a number: 42

C. ID is a number: 42.00

D. Runtime error due to wrong method

Solution

1. Step 1: Analyze type check and method calls

   The function checks if id is a string or number. For number, it calls toFixed(2) which formats number with 2 decimals.

2. Step 2: Apply input and output

   Input is 42 (number), so else branch runs and prints 'ID is a number: 42.00'.

3. Final Answer:

   ID is a number: 42.00 -> Option C

4. Quick Check:

   Number uses toFixed(2) = 42.00 [OK]

Hint: Check typeof before calling type-specific methods [OK]

Common Mistakes:

• Calling string methods on numbers without check
• Expecting no decimal places in toFixed output
• Ignoring type narrowing with typeof

4. Identify the error in this TypeScript code using union types:

function formatInput(input: string | number) {
  return input.toUpperCase();
}

medium

A. No error, code works fine

B. Error: toUpperCase() not available on number

C. Error: Missing return type annotation

D. Error: Union types cannot be used in functions

Solution

1. Step 1: Check method usage on union type

   Method toUpperCase() exists on string but not on number.

2. Step 2: Understand type safety with union

   Without checking type, calling toUpperCase() on input causes error because input might be number.

3. Final Answer:

   Error: toUpperCase() not available on number -> Option B

4. Quick Check:

   Call methods only after type check [OK]

Hint: Always check type before calling type-specific methods [OK]

Common Mistakes:

• Calling string methods on union without type check
• Assuming union types allow all methods
• Ignoring compiler errors about method availability

5. Given this TypeScript code, what will be the output?

type Result = string | null;
function process(value: Result) {
  return value?.toUpperCase() ?? 'No value';
}
console.log(process(null));
console.log(process('hello'));

hard

A. No value\nhello

B. null\nhello

C. Error: Cannot call toUpperCase on null

D. No value\nHELLO

Solution

1. Step 1: Understand optional chaining and nullish coalescing

   The expression value?.toUpperCase() calls toUpperCase() only if value is not null or undefined. Otherwise, it returns undefined.

2. Step 2: Apply nullish coalescing fallback

   If the left side is undefined, ?? 'No value' returns 'No value'. So for null, output is 'No value'. For 'hello', it returns 'HELLO'.

3. Final Answer:

   No value HELLO -> Option D

4. Quick Check:

   Optional chaining + ?? handles null safely [OK]

Hint: Use ?. and ?? to safely handle null in unions [OK]

Common Mistakes:

• Expecting error when calling method on null
• Ignoring optional chaining usage
• Confusing null and undefined handling