Why union types are needed in Typescript - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a variable that can hold either a string or a number.

Typescript

let value: [1];

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

Astring

Bstring & number

Cstring | number

Dnumber

Attempts:

3 left

💡 Hint

Common Mistakes

Using & instead of | which means intersection, not union.

Declaring only one type when both are needed.

The | symbol creates a union type, allowing value to be either a string or a number.

2fill in blank

medium

Complete the function parameter type to accept either a string or an array of strings.

Typescript

function greet(names: [1]) {
  if (typeof names === 'string') {
    console.log(`Hello, ${names}!`);
  } else {
    names.forEach(name => console.log(`Hello, ${name}!`));
  }
}

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

Astring[]

Bany

Cstring & string[]

Dstring | string[]

Attempts:

3 left

💡 Hint

Common Mistakes

Using only string[] which excludes single strings.

Using intersection & which is incorrect here.

The parameter names can be either a single string or an array of strings, so we use a union type string | string[].

3fill in blank

hard

Fix the error in the type annotation to allow the variable to hold a number or null.

Typescript

let result: [1] = null;

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

Anull

Bnumber | null

Cnumber

Dnumber & null

Attempts:

3 left

💡 Hint

Common Mistakes

Using intersection & which causes a type error.

Not including null in the type.

Using | allows result to be either a number or null, which is needed here.

4fill in blank

hard

Fill both blanks to create a function that returns either a string or undefined based on input.

Typescript

function getName(id: number): [1] {
  if (id === 1) {
    return 'Alice';
  }
  return [2];
}

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

Astring | undefined

Bnull

Cundefined

Dstring

Attempts:

3 left

💡 Hint

Common Mistakes

Returning null instead of undefined.

Not using union type for the return type.

The function returns a string or undefined, so the return type is string | undefined and the return value for no match is undefined.

5fill in blank

hard

Fill all three blanks to create a variable that can hold a boolean, number, or string.

Typescript

let data: [1] = true;
data = 42;
data = [2];
data = [3];

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

Aboolean | number | string

B'hello'

C"world"

Dfalse

Attempts:

3 left

💡 Hint

Common Mistakes

Not including all types in the union.

Using wrong quotes for strings.

The variable data can hold boolean, number, or string. The union type is boolean | number | string. The values assigned are true, 42, 'hello', and false.

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