Why union types are needed in Typescript - The Real Reasons

The Big Idea

What if you could write one function that safely handles many types without messy code?

The Scenario

Imagine you are building a form where a user can enter either a phone number or an email address. Without union types, you have to create separate variables or functions for each input type, making your code bulky and confusing.

The Problem

Manually handling multiple possible types means writing repetitive checks and separate code paths. This slows you down and increases the chance of mistakes, like forgetting to handle one type or mixing up data.

The Solution

Union types let you declare a variable or function parameter that can accept multiple types in a clean, simple way. This means you write less code and handle all cases clearly and safely.

Before vs After

✗ Before

function process(input: string) { /* handle string only */ }
function processNumber(input: number) { /* handle number only */ }

✓ After

function process(input: string | number) { /* handle string or number */ }

What It Enables

Union types enable flexible and safe code that can work with different kinds of data without extra complexity.

Real Life Example

When building a chat app, a message might be text or an image URL. Using union types, you can handle both message types in one function easily.

Key Takeaways

Manual handling of multiple types is repetitive and error-prone.

Union types let you combine multiple types in one place.

This leads to cleaner, safer, and more flexible code.

Practice

1. Why do we need union types in TypeScript?
let value: string | number;

easy

A. To make variables immutable

B. To allow a variable to hold more than one type of value

C. To force variables to be only one type

D. To automatically convert types at runtime

Solution

1. Step 1: Understand union type declaration

   The syntax string | number means the variable can hold either a string or a number.

2. Step 2: Identify the purpose of union types

   This flexibility allows one variable to store multiple types safely without errors.

3. Final Answer:

   To allow a variable to hold more than one type of value -> Option B

4. Quick Check:

   Union types = multiple allowed types [OK]

Hint: Union types let variables hold multiple types safely [OK]

Common Mistakes:

• Thinking union types make variables immutable
• Believing union types force a single type only
• Confusing union types with automatic type conversion

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

easy

A. let data: string & number;

B. let data: string or number;

C. let data: string | number;

D. let data: string, number;

Solution

1. Step 1: Recall union type syntax

   Union types use the pipe symbol | to separate types.

2. Step 2: Compare options

   let data: string | number; uses | correctly, while others use invalid syntax.

3. Final Answer:

   let data: string | number; -> Option C

4. Quick Check:

   Union syntax uses | symbol [OK]

Hint: Use | to separate types in union declarations [OK]

Common Mistakes:

• Using & instead of | for union types
• Separating types with commas
• Writing 'or' instead of |

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

function format(value: string | number) {
  if (typeof value === 'string') {
    return value.toUpperCase();
  } else {
    return value.toFixed(2);
  }
}

console.log(format('hello'));
console.log(format(3.14159));

medium

A. "HELLO" and 3

B. "hello" and "3.14159"

C. Error because value can be string or number

D. "HELLO" and "3.14"

Solution

1. Step 1: Analyze function behavior for string input

   If input is a string, toUpperCase() converts it to uppercase, so 'hello' becomes 'HELLO'.

2. Step 2: Analyze function behavior for number input

   If input is a number, toFixed(2) formats it to 2 decimal places, so 3.14159 becomes '3.14'.

3. Final Answer:

   "HELLO" and "3.14" -> Option D

4. Quick Check:

   String upper + number fixed decimals = output [OK]

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

Common Mistakes:

• Assuming no type check needed and causing errors
• Expecting original number without formatting
• Confusing toFixed output as number instead of string

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

function printLength(value: string | number) {
  console.log(value.length);
}
printLength('test');
printLength(123);

medium

A. Error because number type has no 'length' property

B. Error because string type has no 'length' property

C. No error, code runs fine

D. Error because union types cannot be used in functions

Solution

1. Step 1: Understand property access on union types

   Accessing length on value is unsafe because numbers don't have a length property.

2. Step 2: Identify the error cause

   TypeScript will raise an error to prevent runtime failure when value is a number.

3. Final Answer:

   Error because number type has no 'length' property -> Option A

4. Quick Check:

   Accessing length on number causes error [OK]

Hint: Check type before accessing properties on union types [OK]

Common Mistakes:

• Ignoring type checks and expecting no error
• Thinking string has no length property
• Believing union types disallow functions

5. You want a function that accepts either a string or an Array<string> and returns the total number of characters. Which is the best way to use union types to achieve this?

hard

A. function countChars(input: string | string[]) {
if (typeof input === 'string') {
return input.length;
} else {
return input.join('').length;
}
}

B. function countChars(input: string & string[]) {
return input.length;
}

C. function countChars(input: string | string[]) {
return input.length;
}

D. function countChars(input: any) {
return input.length;
}

Solution

1. Step 1: Understand input types and their length

   A string has a length property; an array of strings also has length, but we want total characters, so we join the array first.

2. Step 2: Check union type handling

   function countChars(input: string | string[]) { if (typeof input === 'string') { return input.length; } else { return input.join('').length; } } uses a type check to handle string and array separately, joining array elements to count total characters correctly.

3. Final Answer:

   function countChars(input: string | string[]) { if (typeof input === 'string') { return input.length; } else { return input.join('').length; } } -> Option A

4. Quick Check:

   Use type check and join array for total chars [OK]

Hint: Use typeof and array methods to handle union types properly [OK]

Common Mistakes:

• Using intersection (&) instead of union (|)
• Not checking type before accessing properties
• Using any type losing type safety