With functions, TypeScript infers the types of its parameters as well as any return value.

Function Parameters

The types of function parameters work similarly to variable declarations. If the parameter has a default value, it will take on the type of that value. Otherwise, we may declare the type of that parameter by adding a type annotation after its name.

Here, the logAgeAndName() function explicitly declares age to be of type number, while name is inferred to be of type string from its default value:

function logAgeAndName(age: number, name = '(anonymous)') {
console.log(`${name}, age ${age}`);
logAgeAndName(16, 'Mulan'); // ok: Mulan, age 16
logAgeAndName(0); // also ok: (anonymous), age 0
logAgeAndName('Mulan', 16);
// Argument of type 'string' is not assignable to parameter of type 'number'

The last use of logAgeAndName() shows that the order of the parameters passed to the function matter. The first agrument is expected to be of type number while the second is inferred to be of type string.

Optional Parameters

Function parameters can be made optional by adding a ? question mark after their name, before any type annotation. TypeScript will understand that they don’t need to be provided when the function is invoked. Additionally, their type is a union that includes undefined. This means that if a given function does not use the optional parameter, its value is set to undefined.

The following logFavoriteNumberAndReason() function indicates favorite as a required number and reason as an optional string, so the function must be called with at least one number parameter:

function logFavoriteNumberAndReason(favorite: number, reason?: string) {
console.log(`Favorite: ${favorite}!`);
if (reason) {
console.log(`Because: ${reason}!`);
logFavoriteNumberAndReason(7, 'an esoteric video game reference'); // Ok
logFavoriteNumberAndReason(9001); // Ok
// Error: Expected 1-2 arguments, but got 0.

Return Types

Most functions are written in a way that TypeScript can infer what value they return by looking at all the return statements in the function. Functions that don’t return a value are considered to have a return type of void.

In this example, the getRandomFriend() function is inferred to return a string type because all return statements have a value of type string, including the fallback parameter:

function getRandomFriend(fallback: string) {
switch (Math.floor(Math.random() * 5)) {
case 0:
return 'Josh';
case 1:
return 'Sara';
case 2:
return 'Sonny';
return fallback;
const friend = getRandomFriend('Codey'); // Type: string

The return type of a function can be declared by adding a type annotation after the ) right parenthesis following its list of parameters. Doing so can be useful in two situations:

  • We can make sure the function returns that type.
  • TypeScript will not attempt to infer the return type of a recursive function.

The following recursive fibonacci() function needs an explicit : number return type annotation for TypeScript to understand it. This returns type number:

function fibonacci(i: number): number {
if (i <= 1) {
return i;
return fibonacci(i - 1) + fibonacci(i - 2);

Function Types

Function types may be represented in the type system. They look a lot like an arrow lambda, but with the return type instead of the function body.

This withIncrementedValue() takes in a receiveNewValue parameter function that itself takes in a number and returns nothing (void).

let value = 0;
function withIncrementedValue(receiveNewValue: (newValue: number) => void) {
value += 1;
withIncrementedValue((receivedValue) => {
console.log('Got', receivedValue);

Function parameters’ types may be inferred if their parent function is in a location with a known function type. In the prior example, the receivedValue parameter was inferred to be type number.

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