More on Functions
Functions are the basic building block of any application, whether they're local functions, imported from another module, or methods on a class. They're also values, and just like other values, TypeScript has many ways to describe how functions can be called. Let's learn about how to write types that describe functions.
- Function Type Expressions
- Call Signatures
- Construct Signatures
- Generic Functions
- Inference
- Constraints
- Working with Constrained Values
- Specifying Type Arguments
- Guidelines for Writing Good Generic Functions
- Optional Parameters
- Function Overloads
- Other Types to Know About
- Rest Parameters and Arguments
- Parameter Destructuring
- Assignability of Functions
Function Type Expressions
The simplest way to describe a function is with a a function type expression. These types are syntactically similar to arrow functions:
function greeter(fn: (a: string) => void) { fn("Hello, World"); } function printToConsole(s: string) { console.log(s); } greeter(printToConsole);Try
The syntax (a: string) => void
means "a function with one parameter, named a
, of type string, that doesn't have a return value".
Just like with function declarations, if a parameter type isn't specified, it's implicitly any
.
Note that the parameter name is required. The function type
(string) => void
means "a function with a parameter namedstring
of typea
"!
Of course, we can use a type alias to name a function type:
type GreetFunction = (a: string) => void; function greeter(fn: GreetFunction) { // ... }Try
Call Signatures
In JavaScript, functions can have properties in addition to being callable. However, the function type expression syntax doesn't allow for declaring properties. If we want to describe something callable with properties, we can write a call signature in an object type:
type DescribableFunction = { description: string; (someArg: number): boolean; }; function doSomething(fn: DescribableFunction) { console.log(fn.description + " returned " + fn(6)); }Try
Note that the syntax is slightly different compared to a function type expression - use :
between the parameter list and the return type rather than =>
.
Construct Signatures
JavaScript functions can also be invoked with the new
operator.
TypeScript refers to these as constructors because they usually create a new object.
You can write a construct signature by adding the new
keyword in front of a call signature:
type SomeConstructor = { new(s: string): SomeObject; } function fn(ctor: SomeConstructor) { return new ctor("hello"); }Try
Some objects, like JavaScript's Date
object, can be called with or without new
.
You can combine call and construct signatures in the same type arbitrarily:
interface CallOrConstruct { new(s: string): Date; (n?: number): number; }Try
Generic Functions
It's common to write a function where the types of the input relate to the type of the output, or where the types of two inputs are related in some way. Let's consider for a moment a function that returns the first element of an array:
function firstElement(arr: any[]) { return arr[0]; }Try
This function does its job, but unfortunately has the return type any
.
It'd be better if the function returned the type of the array element.
In TypeScript, generics are used when we want to describe a correspondence between two values. We do this by declaring a type parameter in the function signature:
function firstElement<T>(arr: T[]): T { return arr[0]; }Try
By adding a type parameter T
to this function and using it in two places, we've created a link between the input of the function (the array) and the output (the return value).
Now when we call it, a more specific type comes out:
// s is of type 'string' const s = firstElement(["a", "b", "c"]); // n is of type 'number' const n = firstElement([1, 2, 3]);Try
Inference
Note that we didn't have to specify T
in this sample.
The type was inferred - chosen automatically - by TypeScript.
We can use multiple type parameters as well.
For example, a standalone version of map
would look like this:
function map<E, O>(arr: E[], func: (arg: E) => O): O[] { return arr.map(func); } // Parameter 'n' is of type 'number' // 'parsed' is of type 'string[]' const parsed = map(["1", "2", "3"], n => parseInt(n));Try
Note that in this example, TypeScript could infer both the type of the E
type parameter (from the given string
array), as well as the type O
based on the return value of the function expression.
Constraints
We've written some generic functions that can work on any kind of value. Sometimes we want to relate two values, but can only operate on a certain subset of values. In this case, we can use a constraint to limit the kinds of types that a type parameter can accept.
Let's write a function that returns the longer of two values.
To do this, we need a length
property that's a number.
We constrain the type parameter to that type by writing an extends
clause:
function longest<T extends { length: number }>(a: T, b: T) { if (a.length >= b.length) { return a; } else { return b; } } // longerArray is of type 'number[]' const longerArray = longest([1, 2], [1, 2, 3]); // longerString is of type 'string' const longerString = longest("alice", "bob"); // Error! Numbers don't have a 'length' property const notOK = longest(10Argument of type '10' is not assignable to parameter of type '{ length: number; }'., 100);TryArgument of type '10' is not assignable to parameter of type '{ length: number; }'.
There are a interesting few things to note in this example.
We allowed TypeScript to infer the return type of longest
.
Return type inference also works on generic functions.
Because we constrained T
to { length: number }
, we were allowed to access the .length
property of the a
and b
parameters.
Without the type constraint, we wouldn't be able to access those properties because the values might have been some other type without a length property.
The types of longerArray
and longerString
were inferred based on the arguments.
Remember, generics are all about relating two or more values with the same type!
Finally, just as we'd like, the call to longest(10, 100)
is rejected because the number
type doesn't have a .length
property.
Working with Constrained Values
Here's a common error when working with generic constraints:
function minimumLength<T extends { length: number }>(obj: T, minimum: number): T { if (obj.length >= minimum) { return obj; } else { return { length: minimum };Type '{ length: number; }' is not assignable to type 'T'. } }TryType '{ length: number; }' is not assignable to type 'T'.
It might look like this function is OK - T
is constrained to { length: number }
, and the function either returns T
or a value matching that constraint.
The problem is that the function promises to return the same kind of object as was passed in, not just some object matching the constraint.
If this code were legal, you could write code that definitely wouldn't work:
// 'arr' gets value { length: 6 } const arr = minimumLength([1, 2, 3], 6); // and crashes here because arrays have // a 'slice' method, but not the returned object! console.log(arr.slice(0));Try
Specifying Type Arguments
TypeScript can usually infer the intended type arguments in a generic call, but not always. For example, let's say you wrote a function to combine two arrays:
function combine<T>(arr1: T[], arr2: T[]): T[] { return arr1.concat(arr2); }Try
Normally it would be an error to call this function with mismatched arrays:
const arr = combine([1, 2, 3], ["hello"Type 'string' is not assignable to type 'number'.]);Type 'string' is not assignable to type 'number'.
If you intended to do this, however, you could manually specify T
:
const arr = combine<string | number>([1, 2, 3], ["hello"]);
Guidelines for Writing Good Generic Functions
Writing generic functions is fun, and it can be easy to get carried away with type parameters. Having too many type parameters or using constraints where they aren't needed can make inference less successful, frustrating callers of your function.
Push Type Parameters Down
Here are two ways of writing a function that appear similar:
function firstElement1<T>(arr: T[]) { return arr[0]; } function firstElement2<T extends any[]>(arr: T) { return arr[0]; } // a: number (good) const a = firstElement1([1, 2, 3]); // b: any (bad) const b = firstElement2([1, 2, 3]);Try
These might seem identical at first glance, but firstElement1
is a much better way to write this function.
Its inferred return type is T
, but firstElement2
's inferred return type is any
because TypeScript has to resolve the arr[0]
expression using the constraint type, rather than "waiting" to resolve the element during a call.
Rule: When possible, use the type parameter itself rather than constraining it
Use Fewer Type Parameters
Here's another pair of similar functions:
function filter1<T>(arr: T[], func: (arg: T) => boolean): T[] { return arr.filter(func); } function filter2<T, F extends (arg: T) => boolean>(arr: T[], func: F): T[] { return arr.filter(func); }Try
We've created a type parameter F
that doesn't relate two values.
That's always a red flag, because it means callers wanting to specify type arguments have to manually specify an extra type argument for no reason.
F
doesn't do anything but make the function harder to read and reason about!
Rule: Always use as few type parameters as possible
Type Parameters Should Appear Twice
Sometimes we forget that function doesn't need to be generic:
function greet<S extends string>(s: S) { console.log("Hello, " + s); } greet("world");Try
We could just as easily have written a simpler version:
function greet(s: string) { console.log("Hello, " + s); }Try
Remember, type parameters are for relating the types of multiple values. If a type parameter is only used once in the function signature, it's not relating anything.
Rule: If a type parameter only appears in one location, strongly reconsider if you actually need it
Optional Parameters
Functions in JavaScript often take a variable number of arguments.
For example, the toFixed
method of number
takes an optional digit count:
function f(n: number) { console.log(n.toFixed()); // 0 arguments console.log(n.toFixed(3)); // 1 argument }Try
We can model this in TypeScript by marking the parameter as optional with ?
:
function f(x?: number) { // ... } f(); // OK f(10); // OKTry
Although the parameter is specified as type number
, the x
parameter will actually have the type number | undefined
because unspecified parameters in JavaScript get the value undefined
.
You can also provide a parameter default:
function f(x = 10) { // ... }Try
Now in the body of f
, x
will have type number
because any undefined
argument will be replaced with 10
.
Note that when a parameter is optional, callers can always pass undefined
, as this simply simualtes a "missing" argument:
declare function f(x?: number): void; // cut // All OK f(); f(10); f(undefined);Try
Optional Parameters in Callbacks
Once you've learned about optional parameters and function type expressions, it's very easy to make the following mistakes when writing functions that invoke callbacks:
function myForEach(arr: any[], callback: (arg: any, index?: number) => void) { for (let i = 0; i < arr.length; i++) { callback(arr[i], i); } }Try
What people usually intend when writing index?
as an optional parameter is that they want both of these calls to be legal:
myForEach([1, 2, 3], a => console.log(a)); myForEach([1, 2, 3], (a, i) => console.log(a, i));
What this actually means is that callback
might get invoked with one argument.
In other words, the function definition says that the implementation might look like this:
function myForEach(arr: any[], callback: (arg: any, index?: number) => void) { for (let i = 0; i < arr.length; i++) { // I don't feel like providing the index today callback(arr[i]); } }Try
In turn, TypeScript will enforce this meaning and issue errors that aren't really possible:
myForEach([1, 2, 3], (a, i) => { console.log(iObject is possibly 'undefined'..toFixed()); });TryObject is possibly 'undefined'.
In JavaScript, if you call a function with more arguments than there are parameters, the extra arguments are simply ignored. TypeScript behaves the same way. Functions with fewer parameters (of the same types) can always take the place of functions with more parameters.
When writing a function type for a callback, never write an optional parameter unless you intend to call the function without passing that argument
Function Overloads
Some JavaScript functions can be called in a variety of argument counts and types.
For example, you might write a function to produce a Date
that takes either a timestamp (one argument) or a month/day/year specification (three arguments).
In TypeScript, we can specify a function that can be called in different ways by writing overload signatures. To do this, write some number of function signatures (usually two or more), followed by the body of the function:
function makeDate(timestamp: number): Date; function makeDate(m: number, d: number, y: number): Date; function makeDate(mOrTimestamp: number, d?: number, y?: number): Date { if (d !== undefined && y !== undefined) { return new Date(y, mOrTimestamp, d); } else { return new Date(mOrTimestamp); } } const d1 = makeDate(12345678); const d2 = makeDate(5, 5, 5); const d3 = makeDate(1, 3)No overload expects 2 arguments, but overloads do exist that expect either 1 or 3 arguments.;TryNo overload expects 2 arguments, but overloads do exist that expect either 1 or 3 arguments.
In this example, we wrote two overloads: one accepting one argument, and another accepting three arguments. These first two signatures are called the overload signatures.
Then, we wrote a function implementation with a compatible signature. Functions have an implementation signature, but this signature can't be called directly. Even though we wrote a function with two optional parameters after the required one, it can't be called with two parameters!
Overload Signatures and the Implementation Signature
This is a common source of confusion. Often people will write code like this and not understand why there is an error:
function fn(x: string): void; function fn() { // ... } // Expected to be able to call with zero arguments fn()Expected 1 arguments, but got 0.;TryExpected 1 arguments, but got 0.
Again, the signature used to write the function body can't be "seen" from the outside.
The signature of the implementation is not visible from the outside. When writing an overloaded function, you should always have two or more signatures above the implementation of the function.
The implementation signature must also be compatible with the overload signatures. For example, these functions have errors because the implementation signature doesn't match the overloads in a correct way:
function fn(x: boolean): void; // Argument type isn't right function fnThis overload signature is not compatible with its implementation signature.(x: string): void; function fn(x: boolean) { }TryThis overload signature is not compatible with its implementation signature.
function fn(x: string): string; // Return type isn't right function fnThis overload signature is not compatible with its implementation signature.(x: number): boolean; function fn(x: string | number) { return "oops"; }TryThis overload signature is not compatible with its implementation signature.
Writing Good Overloads
Like generics, there are a few guidelines you should follow when using function overloads. Following these principles will make your function easier to call, easier to understand, and easier to implement.
Let's consider a function that returns the length of a string or an array:
function len(s: string): number; function len(arr: any[]): number; function len(x: any) { return x.length; }Try
This function is fine; we can invoke it with strings or arrays. However, we can't invoke it with a value that might be a string or an array, because TypeScript can only resolve a function call to a single overload:
len(""); // OK len([0]); // OK len(Math.random() > 0.5 ? "hello" : [0]Argument of type 'number[] | "hello"' is not assignable to parameter of type 'any[]'. Type '"hello"' is not assignable to type 'any[]'.);TryArgument of type 'number[] | "hello"' is not assignable to parameter of type 'any[]'.Type '"hello"' is not assignable to type 'any[]'.
Because both overloads have the same argument count and same return type, we can instead write a non-overloaded version of the function:
function len(x: any[] | string) { return x.length; }Try
This is much better! Callers can invoke this with either sort of value, and as an added bonus, we don't have to figure out a correct implementation signature.
Always prefer parameters with union types instead of overloads when possible
Other Types to Know About
There are some additional types you'll want to recognize that appear often when working with function types. Like all types, you can use them everywhere, but these are especially relevant in the context of functions.
void
void
represents the return value of functions which don't return a value.
It's the inferred type any time a function doesn't have any return
statements, or doesn't return any explicit value from those return statements:
// The inferred return type is void function noop() { return; }Try
In JavaScript, a function that doesn't return any value will implicitly return the value undefined
.
However, void
and undefined
are not the same thing in TypeScript.
See the reference page Why void is a special type for a longer discussion about this.
void
is not the same asundefined
.
object
The special type object
refers to any value that isn't a primitive (string
, number
, boolean
, symbol
, null
, or undefined
).
This is different from the empty object type { }
, and also different from the global type Object
.
You can read the reference page about The global types for information on what Object
is for - long story short, don't ever use Object
.
object
is notObject
. Always useobject
!
Note that in JavaScript, function values are objects: They have properties, have Object.prototype
in their prototype chain, are instanceof Object
, you can call Object.keys
on them, and so on.
For this reason, function types are considered to be object
s in TypeScript.
unknown
The unknown
type represents any value.
This is similar to the any
type, but is safer because it's not legal to do anything with an unknown
value:
function f1(a: any) { a.b(); // OK } function f2(a: unknown) { aObject is of type 'unknown'..b(); }TryObject is of type 'unknown'.
This is useful when describing function types because you can describe functions that accept any value without having any
values in your function body.
Conversely, you can describe a function that returns a value of unknown type:
function safeParse(s: string): unknown { return JSON.parse(s); } // Need to be careful with 'obj'! const obj = safeParse(someRandomString);Try
never
Some functions never return a value:
function fail(msg: string): never { throw new Error(msg); }Try
The never
type represents values which are never observed.
In a return type, this means that the function throws an exception or terminates execution of the program.
never
also appears when TypeScript determines there's nothing left in a union.
function fn(x: string | number) { if (typeof x === "string") { // do something } else if (typeof x === "number") { // do something else } else { x; // has type 'never'! } }Try
Function
The global type Function
describes properties like bind
, call
, apply
, and others present on all function values in JavaScript.
It also has the special property that values of type Function
can always be called; these calls return any
:
function doSomething(f: Function) { f(1, 2, 3); }Try
This is an untyped function call and is generally best avoided because of the unsafe any
return type.
If need to accept an arbitrary function but don't intend to call it, the type () => void
is generally safer.
Rest Parameters and Arguments
Background reading: Rest Parameters and Spread Syntax
Rest Parameters
In addition to using optional parameters or overloads to make functions that can accept a variety of fixed argument counts, we can also define functions that take an unbounded number of arguments using rest parameters.
A rest parameter appears after all other parameters, and uses the ...
syntax:
function multiply(n: number, ...m: number[]) { return m.map(x => n * x); } // 'a' gets value [10, 20, 30, 40] const a = multiply(10, 1, 2, 3, 4);Try
In TypeScript, the type annotation on these parameters is implicitly any[]
instead of any
, and any type annotation given must be of the form Array<T>
or T[]
, or a tuple type (which we'll learn about later).
Rest Arguments
Conversely, we can provide a variable number of arguments from an array using the spread syntax.
For example, the push
method of arrays takes any number of arguments:
const arr1 = [1, 2, 3]; const arr2 = [4, 5, 6]; arr1.push(...arr2);Try
Note that in general, TypeScript does not assume that arrays are immutable. This can lead to some surprising behavior:
// Inferred type is number[] -- "an array with zero or more numbers", // not specfically two numbers const args = [8, 5]; const angle = Math.atan2(...argsExpected 2 arguments, but got 0 or more.);TryExpected 2 arguments, but got 0 or more.
The best fix for this situation depends a bit on your code, but in general a const
context is the most straightforward solution:
// Inferred as 2-length tuple const args = [8, 5] as const; // OK const angle = Math.atan2(...args);Try
Parameter Destructuring
You can use parameter destructuring to conveniently unpack objects provided as an argument into one or more local variables in the function body. In JavaScript, it looks like this:
function sum({ a, b, c }) { console.log(a + b + c); } sum({ a: 10, b: 3, c: 9 });Try
The type annotation for the object goes after the destructuring syntax:
function sum({ a, b, c }: { a: number, b: number, c: number }) { console.log(a + b + c); }Try
This can look a bit verbose, but you can use a named type here as well:
// Same as prior example type ABC = { a: number, b: number, c: number }; function sum({ a, b, c }: ABC) { console.log(a + b + c); }Try