TypeScript typeof Function return value

Question

Admit I have a function like this

const createPerson = () => ({ firstName: \'John\', lastName: \'Doe\' })

How can I, without dec

Answer 1

Adapted from https://github.com/Microsoft/TypeScript/issues/14400#issuecomment-291261491

const fakeReturn = <T>(fn: () => T) => ({} as T)

const hello = () => 'World'
const helloReturn = fakeReturn(hello) // {}

type Hello = typeof helloReturn // string

The example in the link uses null as T instead of {} as T, but that breaks with Type 'null' cannot be converted to type 'T'.

The best part is that the function given as parameter to fakeReturn is not actually called.

Tested with TypeScript 2.5.3

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TypeScript 2.8 introduced some predefined conditional types, including the ReturnType<T> that obtains the return type of a function type.

const hello = () => 'World'

type Hello = ReturnType<typeof hello> // string

Answer 2

Original Post

TypeScript < 2.8

I created a little library that permits a workaround, until a fully declarative way is added to TypeScript:

https://npmjs.com/package/returnof

Also created an issue on Github, asking for Generic Types Inference, that would permit a fully declarative way to do this:

https://github.com/Microsoft/TypeScript/issues/14400

---

Update February 2018

TypeScript 2.8

TypeScript 2.8 introduced a new static type ReturnType which permits to achieve that:

https://github.com/Microsoft/TypeScript/pull/21496

You can now easily get the return type of a function in a fully declarative way:

const createPerson = () => ({
  firstName: 'John',
  lastName: 'Doe'
})

type Person = ReturnType<typeof createPerson>

Answer 3

This https://github.com/Microsoft/TypeScript/issues/4233#issuecomment-139978012 might help:

let r = true ? undefined : someFunction();
type ReturnType = typeof r;