In this section, we're going to see how TypeScript can help when a value is one of many possible types. We'll first look at declaring those types using union types, then we'll see how TypeScript can narrow down the type of a value based on your runtime code.
Unions and Literals
Union Types
A union type is TypeScript's way of saying that a value can be "either this type or that type".
This situation comes up in JavaScript all the time. Imagine you have a value that is a string
on Tuesdays, but null
the rest of the time:
const message = Date .now () % 2 === 0 ? 'Hello Tuesdays!' : null
If we hover over message
, we can see that TypeScript has inferred its type as string | null
.
This is a union type. It means that message
can be either a string
or null
.
Declaring Union Types
We can declare our own union types.
For example, you might have an id
that could be either a string
or a number
:
const logId = (id : string | number) => {
console .log (id )
}
This means that logId
can accept either a string
or a number
as an argument, but not a boolean
:
logId ('abc')
logId (123)
logId (true )Argument of type 'boolean' is not assignable to parameter of type 'string | number'.2345Argument of type 'boolean' is not assignable to parameter of type 'string | number'.
To create a union type, the |
operator is used to separate the types. Each type of the union is called a 'member' of the union.
Union types also work when creating your own type aliases. For example, we can refactor our earlier definition into a type alias:
type Id = number | string
function logId(id: Id) {
console.log(id)
}
Union types can contain many different types - they don't all have to be primitives, or don't need to be related in any way. When they get particularly large, you can use this syntax (with the |
before the first member of the union) to make it more readable:
type AllSortsOfStuff =
| string
| number
| boolean
| object
| null
| {
name: string
age: number
}
Union types can be used in many different ways, and they're a powerful tool for creating flexible type definitions.
Literal Types
Just as TypeScript allows us to create union types from multiple types, it also allows us to create types which represent a specific primitive value. These are called literal types.
Literal types can be used to represent strings, numbers, or booleans that have specific values.
type YesOrNo = 'yes' | 'no'
type StatusCode = 200 | 404 | 500
type TrueOrFalse = true | false
In the YesOrNo
type, the |
operator is used to create a union of the string literals "yes"
and "no"
. This means that a value of type YesOrNo
can only be one of these two strings.
This feature is what powers the autocomplete we've seen in functions like document.addEventListener
:
document.addEventListener(
// DOMContentLoaded, mouseover, etc.
'click',
() => {},
)
The first argument to addEventListener
is a union of string literals, which is why we get autocompletion for the different event types.
Combining Unions With Unions
What happens when we make a union of two union types? They combine together to make one big union.
For example, we can create DigitalFormat
and PhysicalFormat
types that contain a union of literal values:
type DigitalFormat = 'MP3' | 'FLAC'
type PhysicalFormat = 'LP' | 'CD' | 'Cassette'
We could then specify AlbumFormat
as a union of DigitalFormat
and `PhysicalFormat:
type AlbumFormat = DigitalFormat | PhysicalFormat
Now, we can use the DigitalFormat
type for functions that handle digital formats, and the AnalogFormat
type for functions that handle analog formats. The AlbumFormat
type can be used for functions that handle all cases.
This way, we can ensure that each function only handles the cases it's supposed to handle, and TypeScript will throw an error if we try to pass an incorrect format to a function.
const getAlbumFormats = (format : PhysicalFormat ) => {
// function body
}
getAlbumFormats ('MP3' )Argument of type '"MP3"' is not assignable to parameter of type 'PhysicalFormat'.2345Argument of type '"MP3"' is not assignable to parameter of type 'PhysicalFormat'.
Exercises
Exercise 1: string
or null
Here we have a function called getUsername
that takes in a username
string. If the username
is not equal to null
, we return a new interpolated string. Otherwise, we return "Guest"
:
function getUsername(username: string) {
if (username !== null) {
return `User: ${username}`
} else {
return 'Guest'
}
}
In the first test, we call getUsername
and pass in a string of "Alice" which passes as expected. However, in the second test, we have a red squiggly line under null
when passing it into getUsername
:
const result = getUsername ('Alice')
type test = Expect <Equal <typeof result , string>>
const result2 = getUsername (null )Argument of type 'null' is not assignable to parameter of type 'string'.2345Argument of type 'null' is not assignable to parameter of type 'string'.
type test2 = Expect <Equal <typeof result2 , string>>
Normally we wouldn't explicitly call the getUsername
function with null
, but in this case it's important that we handle null
values. For example, we might be getting the username
from a user record in a database, and the user might or might not have a name depending on how they signed up.
Currently, the username
parameter only accepts a string
type, and the check for null
isn't doing anything. Update the function parameter's type so the errors are resolved and the function can handle null
.
Exercise 1: string
or null
Exercise 2: Restricting Function Parameters
Here we have a move
function that takes in a direction
of type string, and a distance
of type number:
function move(direction: string, distance: number) {
// Move the specified distance in the given direction
}
The implementation of the function is not important, but the idea is that we want to be able to move either up, down, left, or right.
Here's what calling the move
function might look like:
move('up', 10)
move('left', 5)
To test this function, we have some @ts-expect-error
directives that tell TypeScript we expect the following lines to throw an error.
However, since the move
function currently takes in a string
for the direction
parameter, we can pass in any string we want, even if it's not a valid direction. There is also a test where we expect that passing 20
as a distance won't work, but it's being accepted as well.
This leads to TypeScript drawing red squiggly lines under the @ts-expect-error
directives:
move (
// @ts-expect-error - "up-right" is not a valid direction Unused '@ts-expect-error' directive.2578Unused '@ts-expect-error' directive. 'up-right',
10,
)
move (
// @ts-expect-error - "down-left" is not a valid direction Unused '@ts-expect-error' directive.2578Unused '@ts-expect-error' directive. 'down-left',
20,
)
Your challenge is to update the move
function so that it only accepts the strings "up"
, "down"
, "left"
, and "right"
. This way, TypeScript will throw an error when we try to pass in any other string.
Exercise 2: Restricting Function Parameters
Solution 1: string
or null
The solution is to update the username
parameter to be a union of string
and null
:
function getUsername(username: string | null) {
// function body
}
With this change, the getUsername
function will now accept null
as a valid value for the username
parameter, and the errors will be resolved.
Solution 2: Restricting Function Parameters
In order to restrict what the direction
can be, we can use a union type of literal values (in this case strings).
Here's what this looks like:
function move(direction: 'up' | 'down' | 'left' | 'right', distance: number) {
// Move the specified distance in the given direction
}
With this change, we now have autocomplete for the possible direction
values.
To clean things up a bit, we can create a new type alias called Direction
and update the parameter accordingly:
type Direction = 'up' | 'down' | 'left' | 'right'
function move(direction: Direction, distance: number) {
// Move the specified distance in the given direction
}
Narrowing
Wider vs Narrower Types
Some types are wider versions of other types. For example, string
is wider than the literal string "small"
. This is because string
can be any string, while "small"
can only be the string "small"
.
In reverse, we might say that "small"
is a 'narrower' type than string
. It's a more specific version of a string. 404
is a narrower type than number
, and true
is a narrower type than boolean
.
This is only true of types which have some kind of shared relationship. For example, "small"
is not a narrower version of number
- because "small"
itself is not a number.
In TypeScript, the narrower version of a type can always take the place of the wider version.
For example, if a function accepts a string
, we can pass in "small"
:
const logSize = (size: string) => {
console.log(size.toUpperCase())
}
logSize('small')
But if a function accepts "small"
, we can't pass any random string
:
const recordOfSizes = {
small : 'small',
large : 'large',
}
const logSize = (size : 'small' | 'large') => {
console .log (recordOfSizes [size ])
}
logSize ('medium' )Argument of type '"medium"' is not assignable to parameter of type '"small" | "large"'.2345Argument of type '"medium"' is not assignable to parameter of type '"small" | "large"'.
If you're familiar with the concept of 'subtypes' and 'supertypes' in set theory, this is a similar idea. "small"
is a subtype of string
(it is more specific), and string
is a supertype of "small"
.
Unions Are Wider Than Their Members
A union type is a wider type than its members. For example, string | number
is wider than string
or number
on their own.
This means that we can pass a string
or a number
to a function that accepts string | number
:
function logId(id: string | number) {
console.log(id)
}
logId('abc')
logId(123)
However, this doesn't work in reverse. We can't pass string | number
to a function that only accepts string
.
For example, if we changed this logId
function to only accept a number
, TypeScript would throw an error when we try to pass string | number
to it:
function logId (id : number) {
console .log (`The id is ${id }`)
}
type User = {
id : string | number
}
const user : User = {
id : 123,
}
logId (user . id )Argument of type 'string | number' is not assignable to parameter of type 'number'.
Type 'string' is not assignable to type 'number'.2345Argument of type 'string | number' is not assignable to parameter of type 'number'.
Type 'string' is not assignable to type 'number'.
Hovering over user.id
shows:
Argument of type 'string | number' is not assignable to parameter of type 'number'.
Type 'string' is not assignable to type 'number'.
So, it's important to think of a union type as a wider type than its members.
What is Narrowing?
Narrowing in TypeScript lets us take a wider type and make it narrower using runtime code.
This can be useful when we want to do different things based on the type of a value. For example, we might want to handle a string
differently to a number
, or "small"
differently to "large"
.
Narrowing with typeof
One way we can narrow down the type of a value is to use the typeof
operator, combined with an if
statement.
Consider a function getAlbumYear
that takes in a parameter year
, which can either be a string
or number
. Here's how we could use the typeof
operator to narrow down the type of year
:
const getAlbumYear = (year: string | number) => {
if (typeof year === 'string') {
console.log(`The album was released in ${year.toUppercase()}.`) // `year` is string
} else if (typeof year === 'number') {
console.log(`The album was released in ${year.toFixed(0)}.`) // `year` is number
}
}
It looks straightforward, but there are some important things to realize about what's happening behind the scenes.
Scoping plays a big role in narrowing. In the first if
block, TypeScript understands that year
is a string
because we've used the typeof
operator to check its type. In the else if
block, TypeScript understands that year
is a number
because we've used the typeof
operator to check its type.
This lets us call toUpperCase
on year
when it's a string
, and toFixed
on year
when it's a number
.
However, anywhere outside of the conditional block the type of year
is still the union string | number
. This is because narrowing only applies within the block's scope.
For the sale of illustration, if we add a boolean
to the year
union, the first if
block will still end up with a type of string
, but the else
block will end up with a type of number | boolean
:
const getAlbumYear = (year: string | number | boolean) => {
if (typeof year === 'string') {
console.log(`The album was released in ${year}.`) // `year` is string
} else if (typeof year === 'number') {
console.log(`The album was released in ${year}.`) // `year` is number | boolean
}
console.log(year) // `year` is string | number | boolean
}
This is a powerful example of how TypeScript can read your runtime code and use it to narrow down the type of a value.
Other Ways to Narrow
The typeof
operator is just one way to narrow types.
TypeScript can use other conditional operators like &&
and ||
, and will take the truthiness into account for coercing the boolean value. It's also possible to use other operators like instanceof
and in
for checking object properties. You can even throw errors or use early returns to narrow types.
We'll take a closer look at these in the following exercises.
Exercises
Exercise 1: Narrowing with if
Statements
Here we have a function called validateUsername
that takes in either a string
or null
, and will always return a boolean
:
function validateUsername (username : string | null): boolean {
return username .length > 5'username' is possibly 'null'.18047'username' is possibly 'null'.
return false
}
Tests for checking the length of the username pass as expected:
it('should return true for valid usernames', () => {
expect(validateUsername('Matt1234')).toBe(true)
expect(validateUsername('Alice')).toBe(false)
expect(validateUsername('Bob')).toBe(false)
})
However, we have an error underneath username
inside of the function body, because it could possibly be null
and we are trying to access a property off of it.
it('Should return false for null', () => {
expect(validateUsername(null)).toBe(false)
})
Your task is to rewrite the validateUsername
function to add narrowing so that the null
case is handled and the tests all pass.
Exercise 1: Narrowing with if
Statements
Exercise 2: Throwing Errors to Narrow
Here we have a line of code that uses document.getElementById
to fetch an HTML element, which can return either an HTMLElement
or null
:
const appElement = document.getElementById('app')
Currently, a test to see if the appElement
is an HTMLElement
fails:
type Test = Expect <Equal < typeof appElement , HTMLElement > >Type 'false' does not satisfy the constraint 'true'.2344Type 'false' does not satisfy the constraint 'true'.
Your task is to use throw
to narrow down the type of appElement
before it's checked by the test.
Exercise 2: Throwing Errors to Narrow
Exercise 3: Using in
to Narrow
Here we have a handleResponse
function that takes in a type of APIResponse
, which is a union of two types of objects.
The goal of the handleResponse
function is to check whether the provided object has a data
property. If it does, the function should return the id
property. If not, it should throw an Error
with the message from the error
property.
type APIResponse =
| {
data: {
id: string
}
}
| {
error: string
}
const handleResponse = (response: APIResponse) => {
// How do we check if 'data' is in the response?
if (true) {
return response.data.id
} else {
throw new Error(response.error)
}
}
Currently, there are several errors being thrown as seen in the following tests.
The first error is Property 'data' does not exist on type 'APIResponse'
test('passes the test even with the error', () => {
// there is no error in runtime
expect(() =>
HandleResponseOrThrowError({
error: 'Invalid argument',
}),
).not.toThrowError()
// but the data is returned, instead of an error.
expect(
HandleResponseOrThrowError({
error: 'Invalid argument',
}),
).toEqual("Should this be 'Error'?")
})
Then we have the inverse error, where Property 'error' does not exist on type 'APIResponse'
:
Property data does not exist on type 'APIResponse'.
Your challenge is to find the correct syntax for narrowing down the types within the handleResponse
function's if
condition.
The changes should happen inside of the function without modifying any other parts of the code.
Exercise 3: Using in
to Narrow
Solution 1: Narrowing with if
Statements
There are several different ways to validate the username length.
Option 1: Check Truthiness
We could use an if
statement to check if username
is truthy. If it does, we can return username.length > 5
, otherwise we can return false
:
function validateUsername(username: string | null): boolean {
// Rewrite this function to make the error go away
if (username) {
return username.length > 5
}
return false
}
There is a catch to this piece of logic. If username
is an empty string, it will return false
because an empty string is falsy. This happens to match the behavior we want for this exercise - but it's important to bear that in mind.
Option 2: Check if typeof username
is "string"
We could use typeof
to check if the username is a string:
function validateUsername(username: string | null): boolean {
if (typeof username === 'string') {
return username.length > 5
}
return false
}
This avoids the issue with empty strings.
Option 3: Check if typeof username
is not "string"
Similar to the above, we could check if typeof username !== "string"
.
In this case, if username
is not a string, we know it's null
and could return false
right away. Otherwise, we'd return the check for length being greater than 5:
function validateUsername(username: string | null | undefined): boolean {
if (typeof name !== 'string') {
return false
}
return username.length > 5
}
This shows that TypeScript understands the reverse of a check. Very smart.
Option 4: Check if typeof username
is "object"
A odd JavaScript quirk is that the type of null
is equal to "object"
.
TypeScript knows this, so we can actually use it to our advantage. We can check if username
is an object, and if it is, we can return false
:
function validateUsername(username: string | null): boolean {
if (typeof username === 'object') {
return false
}
return username.length > 5
}
Option 5: Extract the check into its own variable
Finally, for readability and reusability purposes you could store the check in its own variable isUsernameOK
.
Here's what this would look like:
function validateUsername(username: string | null): boolean {
const isUsernameOK = typeof username === 'string'
if (isUsernameOK) {
return username.length > 5
}
return false
}
TypeScript is smart enough to understand that the value of isUsernameOK
corresponds to whether username
is a string or not. Very smart.
All of the above options use if
statements to perform checks by narrowing types by using typeof
.
No matter which option you go with, remember that you can always use an if
statement to narrow your type and add code to the case that the condition passes.
Solution 2: Throwing Errors to Narrow
The issue with this code is that document.getElementById
returns null | HTMLElement
. But we want to make sure that appElement
is an HTMLElement
before we use it.
We are pretty sure that appElement
exists. If it doesn't exist, we probably want to crash the app early so that we can get an informative error about what's gone wrong.
So, we can add an if
statement that checks if appElement
is falsy, then throws an error:
if (!appElement) {
throw new Error('Could not find app element')
}
By adding this error condition, we can be sure that we will never reach any subsequent code if appElement
is null
.
If we hover over appElement
after the if
statement, we can see that TypeScript now knows that appElement
is an HTMLElement
- it's no longer null
. This means our test also now passes:
console .log (appElement )
type Test = Expect <Equal <typeof appElement , HTMLElement >> // passes
Throwing errors like this can help you identify issues at runtime. In this specific case, it narrows down the code outside of the immediate if
statement scope. Amazing.
Solution 3: Using in
to Narrow
Your first instinct will be to check if response.data
is truthy.
const handleResponse = (response : APIResponse ) => {
if (response .data ) {Property 'data' does not exist on type 'APIResponse'.
Property 'data' does not exist on type '{ error: string; }'.2339Property 'data' does not exist on type 'APIResponse'.
Property 'data' does not exist on type '{ error: string; }'. return response .data .id Property 'data' does not exist on type 'APIResponse'.
Property 'data' does not exist on type '{ error: string; }'.2339Property 'data' does not exist on type 'APIResponse'.
Property 'data' does not exist on type '{ error: string; }'. } else {
throw new Error (response .error )Property 'error' does not exist on type 'APIResponse'.
Property 'error' does not exist on type '{ data: { id: string; }; }'.2339Property 'error' does not exist on type 'APIResponse'.
Property 'error' does not exist on type '{ data: { id: string; }; }'. }
}
But you'll get an error. This is because response.data
is only available on one of the members of the union. TypeScript doesn't know that response
is the one with data
on it.
Option 1: Changing the Type
It may be tempting to change the APIResponse
type to add .data
to both branches:
type APIResponse =
| {
data: {
id: string
}
}
| {
data?: undefined
error: string
}
This is certainly one way to handle it. But there is a built-in way to do it.
Option 2: Using in
We can use an in
operator to check if a specific key exists on response
.
In this example, it would check for the key data
:
const handleResponse = (response : APIResponse ) => {
if ('data' in response ) {
return response .data .id
} else {
throw new Error (response .error )
}
}
If the response
isn't the one with data
on it, then it must be the one with error
, so we can throw an Error
with the error message.
You can check this out by hovering over .data
and .error
in each of the branches of the if
statement. TypeScript will show you that it knows the type of response
in each case.
Using in
here gives us a great way to narrow down objects that might have different keys from one another.
unknown
and never
Let's pause for a moment to introduce a couple more types that play an important role in TypeScript, particularly when we talk about 'wide' and 'narrow' types.
The Widest Type: unknown
TypeScript's widest type is unknown
. It represents something that we don't know what it is.
If you imagine a scale whether the widest types are at the top and the narrowest types are at the bottom, unknown
is at the top. All other types like strings, numbers, booleans, null, undefined, and their respective literals are assignable to unknown
, as seen in its assignability chart:
Consider this example function fn
that takes in an input
parameter of type unknown
:
const fn = (input : unknown) => {}
// Anything is assignable to unknown!
fn ('hello')
fn (42)
fn (true)
fn ({})
fn ([])
fn (() => {})
All of the above function calls are valid because unknown
is assignable to any other type
The unknown
type is the preferred choice when you want to represent something that's truly unknown in JavaScript. For example, it is extremely useful when you have things coming into your application from outside sources, like input from a form or a call to a webhook.
What's the Difference Between unknown
and any
?
You might be wondering what the difference is between unknown
and any
. They're both wide types, but there's a key difference.
any
doesn't really fit into our definition of 'wide' and 'narrow' types. It breaks the type system. It's not really a type at all - it's a way of opting out of TypeScript's type checking.
any
can be assigned to anything, and anything can be assigned to any
. any
is both narrower and wider than every other type.
unknown
, on the other hand, is part of TypeScript's type system. It's wider than every other type, so it can't be assigned to anything.
const handleWebhookInput = (input : unknown) => {
input .toUppercase ()'input' is of type 'unknown'.18046'input' is of type 'unknown'.}
const handleWebhookInputWithAny = (input : any) => {
// no error
input .toUppercase ()
}
This means that unknown
is a safe type, but any
is not. unknown
means "I don't know what this is", while any
means "I don't care what this is".
The Narrowest Type: never
If unknown
is the widest type in TypeScript, never
is the narrowest.
never
represents something that will never happen. It's the very bottom of the type hierarchy.
You'll rarely use a never
type annotation yourself. Instead, it'll pop up in error messages and hovers - often when narrowing.
But first, let's look at a simple example of a never
type:
never
vs void
Let's consider a function that never returns anything:
const getNever = () => {
// This function never returns!
}
When hovering this function, TypeScript will infer that it returns void
, indicating that it essentially returns nothing.
// hovering over `getNever` shows:
const getNever: () => void
However, if we throw an error inside of the function, the function will never return:
const getNever = () => {
throw new Error('This function never returns')
}
With this change, TypeScript will infer that the function's type is never
:
// hovering over `getNever` shows:
const getNever: () => never
The never
type represents something that can never happen.
There are some weird implications for the never
type.
You cannot assign anything to never
, except for never
itself.
const fn = (input : never) => {}
fn ('hello' )Argument of type 'string' is not assignable to parameter of type 'never'.2345Argument of type 'string' is not assignable to parameter of type 'never'.fn (42 )Argument of type 'number' is not assignable to parameter of type 'never'.2345Argument of type 'number' is not assignable to parameter of type 'never'.fn (true )Argument of type 'boolean' is not assignable to parameter of type 'never'.2345Argument of type 'boolean' is not assignable to parameter of type 'never'.fn ({} )Argument of type '{}' is not assignable to parameter of type 'never'.2345Argument of type '{}' is not assignable to parameter of type 'never'.fn ([] )Argument of type 'never[]' is not assignable to parameter of type 'never'.2345Argument of type 'never[]' is not assignable to parameter of type 'never'.fn (() => {} )Argument of type '() => void' is not assignable to parameter of type 'never'.2345Argument of type '() => void' is not assignable to parameter of type 'never'.
// no error here, since we're assigning `never` to `never`
fn (getNever ())
However, you can assign never
to anything:
const str: string = getNever()
const num: number = getNever()
const bool: boolean = getNever()
const arr: string[] = getNever()
This behavior looks extremely odd at first - but we'll see later why it's useful.
Let's update our chart to include never
:
This gives us pretty much the full picture of TypeScript's type hierarchy.
Exercises
Exercise 1: Narrowing Errors with instanceof
In TypeScript, one of the most common places you'll encounter the unknown
type is when using a try...catch
statement to handle potentially dangerous code. Let's consider an example:
const somethingDangerous = () => {
if (Math .random () > 0.5) {
throw new Error ('Something went wrong')
}
return 'all good'
}
try {
somethingDangerous ()
} catch (error ) {
if (true) {
console .error (error .message )'error' is of type 'unknown'.18046'error' is of type 'unknown'. }
}
In the code snippet above, we have a function called somethingDangerous
that has a 50/50 chance of either throwing an error.
Notice that the error
variable in the catch
clause is typed as unknown
.
Now let's say we want to log the error using console.error()
only if the error contains a message
attribute. We know that errors typically come with a message
attribute, like in the following example:
const error = new Error('Some error message')
console.log(error.message)
Your task is to update the if
statement to have the proper condition to check if the error
has a message attribute before logging it. Check the title of the exercise to get a hint... And remember, Error
is a class.
Exercise 1: Narrowing Errors with instanceof
Exercise 2: Narrowing unknown
to a Value
Here we have a parseValue
function that takes in a value
of type unknown
:
const parseValue = (value : unknown) => {
if (true) {
return value .data .id 'value' is of type 'unknown'.18046'value' is of type 'unknown'. }
throw new Error ('Parsing error!')
}
The goal of this function is to return the id
property of the data
property of the value
object. If the value
object doesn't have a data
property, then it should throw an error.
Here are some tests for the function that show us the amount of narrowing that needs to be done inside of the parseValue
function:
it('Should handle a { data: { id: string } }', () => {
const result = parseValue({
data: {
id: '123',
},
})
type test = Expect<Equal<typeof result, string>>
expect(result).toBe('123')
})
it('Should error when anything else is passed in', () => {
expect(() => parseValue('123')).toThrow('Parsing error!')
expect(() => parseValue(123)).toThrow('Parsing error!')
})
Your challenge is to modify the parseValue
function so that the tests pass and the errors go away. I want you to challenge yourself to do this only by narrowing the type of value
inside of the function. No changes to the types. This will require a very large if
statement!
Exercise 2: Narrowing unknown
to a Value
Exercise 3: Reusable Type Guards
Let's imagine that we have two very similar functions, each with a long conditional check to narrow down the type of a value.
Here's the first function:
const parseValue = (value: unknown) => {
if (
typeof value === 'object' &&
value !== null &&
'data' in value &&
typeof value.data === 'object' &&
value.data !== null &&
'id' in value.data &&
typeof value.data.id === 'string'
) {
return value.data.id
}
throw new Error('Parsing error!')
}
And here's the second function:
const parseValueAgain = (value: unknown) => {
if (
typeof value === 'object' &&
value !== null &&
'data' in value &&
typeof value.data === 'object' &&
value.data !== null &&
'id' in value.data &&
typeof value.data.id === 'string'
) {
return value.data.id
}
throw new Error('Parsing error!')
}
Both functions have the same conditional check. This is a great opportunity to create a reusable type guard.
All the tests are currently passing. Your job is to try to refactor the two functions to use a reusable type guard, and remove the duplicated code. As it turns out, TypeScript makes this a lot easier than you expect.
Exercise 3: Reusable Type Guards
Solution 1: Narrowing Errors with instanceof
The way to solve this challenge is to narrow the error
using the instanceof
operator.
Where we check the error message, we'll check if error
is an instance of the class Error
:
if (error instanceof Error) {
console.log(error.message)
}
The instanceof
operator covers other classes which inherit from the Error
class as well, such as TypeError
.
In this case, we're logging the error message to the console - but this could be used to display something different in our applications, or to log the error to an external service.
Even though it works in this particular example for all kinds of Error
s, it won't cover us for the strange case where someone throws a non-Error
object.
throw 'This is not an error!'
To be more safe from these edge cases, it's a good idea to include an else
block that would throw the error
variable like so:
if (error instanceof Error) {
console.log(error.message)
} else {
throw error
}
Using this technique, we can handle the error in a safe way and avoid any potential runtime errors.
Solution 2: Narrowing unknown
to a Value
Here's our starting point:
const parseValue = (value : unknown) => {
if (true) {
return value .data .id 'value' is of type 'unknown'.18046'value' is of type 'unknown'. }
throw new Error ('Parsing error!')
}
To fix the error, we'll need to narrow the type using conditional checks. Let's take it step-by-step.
First, we'll check if the type of value
is an object
by replacing the true
with a type check:
const parseValue = (value : unknown) => {
if (typeof value === 'object') {
return value .data .id Property 'data' does not exist on type 'object'.2339Property 'data' does not exist on type 'object'.'value' is possibly 'null'.18047'value' is possibly 'null'. }
throw new Error ('Parsing error!')
}
Then we'll check if the value
argument has a data
attribute using the in
operator:
const parseValue = (value : unknown) => {
if (typeof value === 'object' && 'data' in value ) {'value' is possibly 'null'.18047'value' is possibly 'null'. return value . data .id 'value.data' is of type 'unknown'.18046'value.data' is of type 'unknown'. }
throw new Error ('Parsing error!')
}
With this change, TypeScript is complaining that value
is possibly null
. This is because, of course, typeof null
is "object"
. Thanks, JavaScript!
To fix this, we can add && value
to our first condition to make sure it isn't null
:
const parseValue = (value : unknown) => {
if (typeof value === 'object' && value && 'data' in value ) {
return value . data .id 'value.data' is of type 'unknown'.18046'value.data' is of type 'unknown'. }
throw new Error ('Parsing error!')
}
Now our condition check is passing, but we're still getting an error on value.data
being typed as unknown
.
What we need to do now is to narrow the type of value.data
to an object
and make sure that it isn't null
. At this point we'll also add specify a return type of string
to avoid returning an unknown
type:
const parseValue = (value : unknown): string => {
if (
typeof value === 'object' &&
value !== null &&
'data' in value &&
typeof value .data === 'object' &&
value .data !== null
) {
return value .data .id Property 'id' does not exist on type 'object'.2339Property 'id' does not exist on type 'object'. }
throw new Error ('Parsing error!')
}
Finally, we'll add a check to ensure that the id
is a string. If not, TypeScript will throw an error:
const parseValue = (value: unknown): string => {
if (
typeof value === 'object' &&
value !== null &&
'data' in value &&
typeof value.data === 'object' &&
value.data !== null &&
'id' in value.data &&
typeof value.data.id === 'string'
) {
return value.data.id
}
throw new Error('Parsing error!')
}
Now when we hover over parseValue
, we can see that it takes in an unknown
input and always returns a string
:
// hovering over `parseValue` shows:
const parseValue: (value: unknown) => string
Thanks to this huge conditional, our tests pass, and our error messages are gone!
This is usually not how you'd want to write your code. It's a bit of a mess. You could use a library like Zod to do this with a much nicer API. But it's a great way to understand how unknown
and narrowing work in TypeScript.
Solution 3: Reusable Type Guards
The first step is to create a function called hasDataId
that captures the conditional check:
const hasDataId = (value) => {
return (
typeof value === 'object' &&
value !== null &&
'data' in value &&
typeof value.data === 'object' &&
value.data !== null &&
'id' in value.data &&
typeof value.data.id === 'string'
)
}
We haven't given value
a type here - unknown
makes sense, because it could be anything.
Now we can refactor the two functions to use this type guard:
const parseValue = (value: unknown) => {
if (hasDataId(value)) {
return value.data.id
}
throw new Error('Parsing error!')
}
const parseValueAgain = (value: unknown) => {
if (hasDataId(value)) {
return value.data.id
}
throw new Error('Parsing error!')
}
Incredibly, this is all TypeScript needs to be able to narrow the type of value
inside of the if
statement. It's smart enough to understand that hasDataId
being called on value
ensures that value
has a data
property with an id
property.
We can observe this by hovering over hasDataId
:
// hovering over `hasDataId` shows:
const hasDataId: (value: unknown) => value is {data: {id: string}}
This return type we're seeing is a type predicate. It's a way of saying "if this function returns true
, then the type of the value is { data: { id: string } }
".
We'll look at authoring our own type predicates in one of the later chapters in the book - but it's very useful that TypeScript infers its own.
Discriminated Unions
In this section we'll look at a common pattern TypeScript developers use to structure their code. It's called a 'discriminated union'.
To understand what a discriminated union is, let's first look at the problem it solves.
The Problem: The Bag Of Optionals
Let's imagine we are modelling a data fetch. We have a State
type with a status
property which can be in one of three states: loading
, success
, or error
.
type State = {
status: 'loading' | 'success' | 'error'
}
This is useful, but we also need to capture some extra data. The data coming back from the fetch, or the error message if the fetch fails.
We could add an error
and data
property to the State
type:
type State = {
status: 'loading' | 'success' | 'error'
error?: string
data?: string
}
And let's imagine we have a renderUI
function that returns a string based on the input.
const renderUI = (state : State ) => {
if (state .status === 'loading') {
return 'Loading...'
}
if (state .status === 'error') {
return `Error: ${state . error .toUpperCase ()}`'state.error' is possibly 'undefined'.18048'state.error' is possibly 'undefined'. }
if (state .status === 'success') {
return `Data: ${state .data }`
}
}
This all looks good, except for the error we're getting on state.error
. TypeScript is telling us that state.error
could be undefined
, and we can't call toUpperCase
on undefined
.
This is because we've declared our State
type in an incorrect way. We've made it so the error
and data
properties are not related to the statuses where they occur. In other words, it's possible to create types which will never happen in our app:
const state: State = {
status: 'loading',
error: 'This is an error', // should not happen on "loading!"
data: 'This is data', // should not happen on "loading!"
}
I'd describe this type as a "bag of optionals". It's a type that's too loose. We need to tighten it up so that error
can only happen on error
, and data
can only happen on success
.
The Solution: Discriminated Unions
The solution is to turn our State
type into a discriminated union.
A discriminated union is a type that has a common property, the 'discriminant', which is a literal type that is unique to each member of the union.
In our case, the status
property is the discriminant.
Let's take each status and separate them into separate object literals:
type State =
| {
status: 'loading'
}
| {
status: 'error'
}
| {
status: 'success'
}
Now, we can associate the error
and data
properties with the error
and success
statuses respectively:
type State =
| {
status: 'loading'
}
| {
status: 'error'
error: string
}
| {
status: 'success'
data: string
}
Now, if we hover over state.error
in the renderUI
function, we can see that TypeScript knows that state.error
is a string
:
const renderUI = (state : State ) => {
if (state .status === 'loading') {
return 'Loading...'
}
if (state .status === 'error') {
return `Error: ${state .error .toUpperCase ()}` }
if (state .status === 'success') {
return `Data: ${state .data }`
}
}
This is due to TypeScript's narrowing - it knows that state.status
is "error"
, so it knows that state.error
is a string
inside of the if
block.
To clean up our original type, we could use a type alias for each of the statuses:
type LoadingState = {
status: 'loading'
}
type ErrorState = {
status: 'error'
error: string
}
type SuccessState = {
status: 'success'
data: string
}
type State = LoadingState | ErrorState | SuccessState
So if you're noticing that your types are resembling 'bags of optionals', it's a good idea to consider using a discriminated union.
Exercises
Exercise 1: Destructuring a Discriminated Union
Consider a discriminated union called Shape
that is made up of two types: Circle
and Square
. Both types have a kind
property that acts as the discriminant.
type Circle = {
kind: 'circle'
radius: number
}
type Square = {
kind: 'square'
sideLength: number
}
type Shape = Circle | Square
This calculateArea
function destructures the kind
, radius
, and sideLength
properties from the Shape
that is passed in, and calculates the area of the shape accordingly:
function calculateArea ({kind , radius , sideLength }: Shape ) {Property 'sideLength' does not exist on type 'Shape'.2339Property 'sideLength' does not exist on type 'Shape'.Property 'radius' does not exist on type 'Shape'.2339Property 'radius' does not exist on type 'Shape'. if (kind === 'circle') {
return Math .PI * radius * radius
} else {
return sideLength * sideLength
}
}
However, TypeScript is showing us errors below 'radius'
and 'sideLength'
.
Your task is to update the implementation of the calculateArea
function so that destructuring properties from the passed in Shape
works without errors. Hint: the examples I showed in the chapter didn't use destructuring, but some destructuring is possible.
Exercise 1: Destructuring a Discriminated Union
Exercise 2: Narrowing a Discriminated Union with a Switch Statement
Here we have our calculateArea
function from the previous exercise, but without any destructuring.
function calculateArea(shape: Shape) {
if (shape.kind === 'circle') {
return Math.PI * shape.radius * shape.radius
} else {
return shape.sideLength * shape.sideLength
}
}
Your challenge is to refactor this function to use a switch
statement instead of the if/else
statement. The switch
statement should be used to narrow the type of shape
and calculate the area accordingly.
Exercise 2: Narrowing a Discriminated Union with a Switch Statement
Exercise 3: Discriminated Tuples
Here we have a fetchData
function that returns a promise that resolves to an APIResponse
tuple that consists of two elements.
The first element is a string that indicates the type of the response. The second element can be either an array of User
objects in the case of successful data retrieval, or a string in the event of an error:
type APIResponse = [string, User[] | string]
Here's what the fetchData
function looks like:
async function fetchData(): Promise<APIResponse> {
try {
const response = await fetch('https://api.example.com/data')
if (!response.ok) {
return [
'error',
// Imagine some improved error handling here
'An error occurred',
]
}
const data = await response.json()
return ['success', data]
} catch (error) {
return ['error', 'An error occurred']
}
}
However, as seen in the tests below, the APIResponse
type currently will allow for other combinations that aren't what we want. For example, it would allow for passing an error message when data is being returned:
async function exampleFunc () {
const [status , value ] = await fetchData ()
if (status === 'success') {
console .log (value )
type test = Expect <Equal < typeof value , User []> >Type 'false' does not satisfy the constraint 'true'.2344Type 'false' does not satisfy the constraint 'true'. } else {
console .error (value )
type test = Expect <Equal < typeof value , string > >Type 'false' does not satisfy the constraint 'true'.2344Type 'false' does not satisfy the constraint 'true'. }
}
The problem stems from the APIResponse
type being a "bag of optionals".
The APIResponse
type needs to be updated so that there are two possible combinations for the returned tuple:
If the first element is "error"
then the second element should be the error message.
If the first element is "success"
, then the second element should be an array of User
objects.
Your challenge is to redefine the APIResponse
type to be a discriminated tuple that only allows for the specific combinations for the success
and error
states defined above.
Exercise 3: Discriminated Tuples
Exercise 4: Handling Defaults with a Discriminated Union
We're back with our calculateArea
function:
function calculateArea(shape: Shape) {
if (shape.kind === 'circle') {
return Math.PI * shape.radius * shape.radius
} else {
return shape.sideLength * shape.sideLength
}
}
Until now, the test cases have involved checking if the kind
of the Shape
is a circle
or a square
, then calculating the area accordingly.
However, a new test case has been added for a situation where no kind
has been passed into the function:
// @errors: 2345
import {Equal, Expect} from '@total-typescript/helpers'
import {it, expect} from 'vitest'
type Circle = {
kind: 'circle'
radius: number
}
type Square = {
kind: 'square'
sideLength: number
}
type Shape = Circle | Square
function calculateArea(shape: Shape) {
if (shape.kind === 'circle') {
return Math.PI * shape.radius * shape.radius
} else {
return shape.sideLength * shape.sideLength
}
}
// ---cut---
it('Should calculate the area of a circle when no kind is passed', () => {
const result = calculateArea({
radius: 5,
})
expect(result).toBe(78.53981633974483)
type test = Expect<Equal<typeof result, number>>
})
TypeScript is showing errors under radius
in the test:
The test expects that if a kind
isn't passed in, the shape should be treated as a circle. However, the current implementation doesn't account for this.
Your challenge is to:
- Make updates to the
Shape
discriminated union that will allow for us to omitkind
. - Make adjustments to the
calculateArea
function to ensure that TypeScript's type narrowing works properly within the function.
Exercise 4: Handling Defaults with a Discriminated Union
Solution 1: Destructuring a Discriminated Union
Before we look at the working solution, let's look at an attempt that doesn't work out.
A Non-Working Attempt at Destructuring Parameters
Since we know that kind
is present in all branches of the discriminated union, we can try using the rest parameter syntax to bring along the other properties:
function calculateArea({kind, ...shape}: Shape) {
// rest of function
}
Then inside of the conditional branches, we can specify the kind
and destructure from the shape
object:
function calculateArea ({kind , ...shape }: Shape ) {
if (kind === 'circle') {
const {radius } = shape Property 'radius' does not exist on type '{ radius: number; } | { sideLength: number; }'.2339Property 'radius' does not exist on type '{ radius: number; } | { sideLength: number; }'.
return Math .PI * radius * radius
} else {
const {sideLength } = shape Property 'sideLength' does not exist on type '{ radius: number; } | { sideLength: number; }'.2339Property 'sideLength' does not exist on type '{ radius: number; } | { sideLength: number; }'.
return sideLength * sideLength
}
}
However, this approach doesn't work because the kind
property has been separated from the rest of the shape. As a result, TypeScript can't track the relationship between kind
and the other properties of shape
. Both radius
and sideLength
have error messages below them.
TypeScript gives us these errors because it still cannot guarantee properties in the function parameters since it doesn't know yet whether it's dealing with a Circle
or a Square
.
The Working Destructuring Solution
Instead of doing the destructuring at the function parameter level, we instead will revert the function parameter back to shape
:
function calculateArea(shape: Shape) {
// rest of function
}
...and move the destructuring to take place inside of the conditional branches:
function calculateArea(shape: Shape) {
if (shape.kind === 'circle') {
const {radius} = shape
return Math.PI * radius * radius
} else {
const {sideLength} = shape
return sideLength * sideLength
}
}
Now within the if
condition, TypeScript can recognize that shape
is indeed a Circle
and allows us to safely access the radius
property. A similar approach is taken for the Square
in the else
condition.
This approach works because TypeScript can track the relationship between kind
and the other properties of shape
when the destructuring takes place inside of the conditional branches.
In general, I prefer to avoid destructuring when working with discriminated unions. But if you want to, do it inside of the conditional branches.
Solution 2: Narrowing a Discriminated Union with a Switch Statement
The first step is to clear out the calculateArea
function and add the switch
keyword and specify shape.kind
as our switch condition:
function calculateArea(shape: Shape) {
switch (shape.kind) {
case 'circle': {
return Math.PI * shape.radius * shape.radius
}
case 'square': {
return shape.sideLength * shape.sideLength
}
// Potential additional cases for more shapes
}
}
As a nice bonus, TypeScript offers us autocomplete on the cases for the switch
statement. This is a great way to ensure that we're handling all of the cases for our discriminated union.
Not Accounting for All Cases
As an experiment, comment out the case where the kind
is square
:
function calculateArea(shape: Shape) {
switch (shape.kind) {
case 'circle': {
return Math.PI * shape.radius * shape.radius
}
// case "square": {
// return shape.sideLength * shape.sideLength;
// }
// Potential additional cases for more shapes
}
}
Now when we hover over the function, we see that the return type is number | undefined
. This is because TypeScript is smart enough to know that if we don't return a value for the square
case, the output will be undefined
for any square
shape.
// hovering over `calculateArea` shows
function calculateArea(shape: Shape): number | undefined
Switch statements work great with discriminated unions!
Solution 3: Destructuring a Discriminated Union of Tuples
When you're done, your APIResponse
type should look like this:
type APIResponse = ['error', string] | ['success', User[]]
We've created two possible combinations for the APIResponse
type. An error state, and a success state. And instead of objects, we've used tuples.
You might be thinking - where's the discriminant? It's the first element of the tuple. This is what's called a discriminated tuple.
And with this update to the APIResponse
type, the errors have gone away!
Understanding Tuple Relationships
Inside of the exampleFunc
function, we use array destructuring to pull out the status
and value
from the APIResponse
tuple:
const [status, value] = await fetchData()
Even though the status
and value
variables are separate, TypeScript keeps track of the relationships behind them. If status
is checked and is equal to "success"
, TypeScript can narrow down value
to be of the User[]
type automatically:
// hovering over `status` shows
const status: 'error' | 'success'
Note that this intelligent behavior is specific to discriminated tuples, and won't work with discriminated objects - as we saw in our previous exercise.
Solution 4: Handling Defaults with a Discriminated Union
Before we look at the working solution, let's take a look at a couple of approaches that don't quite work out.
Attempt 1: Creating an OptionalCircle
Type
One possible first step is to create an OptionalCircle
type by discarding the kind
property:
type OptionalCircle = {
radius: number
}
Then we would update the Shape
type to include the new type:
type Shape = Circle | OptionalCircle | Square
This solution appears to work initially since it resolves the error in the radius test case.
However, this approach brings back errors inside of the calculateArea
function because the discriminated union is broken since not every member has a kind
property.
function calculateArea(shape: Shape) {
if (shape.kind === 'circle') {
// error on shape.kind
return Math.PI * shape.radius * shape.radius
} else {
return shape.sideLength * shape.sideLength
}
}
Attempt 2: Updating the Circle
Type
Rather than developing a new type, we could modify the Circle
type to make the kind
property optional:
type Circle = {
kind?: 'circle'
radius: number
}
type Square = {
kind: 'square'
sideLength: number
}
type Shape = Circle | Square
This modification allows us to distinguish between circles and squares. The discriminated union remains intact while also accommodating the optional case where kind
is not specified.
However, there is now a new error inside of the calculateArea
function:
function calculateArea (shape : Shape ) {
if (shape .kind === 'circle') {
return Math .PI * shape .radius * shape .radius
} else {
return shape .sideLength * shape .sideLength Property 'sideLength' does not exist on type 'Shape'.
Property 'sideLength' does not exist on type 'Circle'.2339Property 'sideLength' does not exist on type 'Shape'.
Property 'sideLength' does not exist on type 'Circle'.Property 'sideLength' does not exist on type 'Shape'.
Property 'sideLength' does not exist on type 'Circle'.2339Property 'sideLength' does not exist on type 'Shape'.
Property 'sideLength' does not exist on type 'Circle'. }
}
The error tells us that TypeScript is no longer able to narrow down the type of shape
to a Square
because we're not checking to see if shape.kind
is undefined
.
Fixing the New Error
It would be possible to fix this error by adding additional checks for the kind
, but instead we could just swap how our conditional checks work.
We'll check for a square
first, then fall back to a circle
:
if (shape.kind === 'square') {
return shape.sideLength * shape.sideLength
} else {
return Math.PI * shape.radius * shape.radius
}
By inspecting square
first, all shape cases that aren't squares default to circles. The circle is treated as optional, which preserves our discriminated union and keeps the function flexible.
Sometimes, just flipping the runtime logic makes TypeScript happy!