Declare a Typescript function that complies to a type

Question

Let's suppose I have declared a type containing defining a specific type of function

type Callback = (err: Error | null, result: any)
type UselessFunction = (event: string, context: any, callback: Callback) => void

I would like to declare functions that must comply to the type UselessFunction

I only found in the documentation how to assign a type to a function using arrow syntax

const someFunction: UselessFunction = (evt, ctx, cb) => {}

How can a type be assigned be function declaration? I cannot find the syntax in the docs

function someFunction (event, context, cb) {
}

If you type the individual parameters of someFunction, it should gain the type UselessFunction where necessary by duck typing — MTCoster
– MTCoster, Commented Feb 14, 2019 at 16:19

Mateusz Kocz · Accepted Answer · 2019-02-14 18:06:06Z

3

TypeScript lacks this functionality. You need to add the type annotation to the declaration's parameters and return value separately to make this function be of a specific type. You can, however, make it a bit more convenient by using Parameters and ReturnType types.

type UselessFunction = (event: string, context: any, callback: Callback) => void

function someFunction (...params: Parameters<UselessFunction>): ReturnType<UselessFunction> {
  // ...
}

answered Feb 14, 2019 at 18:06

Mateusz Kocz

4,6021 gold badge27 silver badges27 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

Declare a Typescript function that complies to a type

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related