如何在 TypeScript 中导入导出函数的派生类型

Question

I have two files myFunction.ts and index.ts .

• myFunction.ts exports a function like so:

export default (param1: string) => { return true }

• index.ts imports that function like so:

import myFunction from './myFunction'

and then I want to use the typings from myFunction like so:

function bla(aFn: myFunction) {
  aFn('hello')
}

However, the compiler gives me a cannot find name myFunction .

How do I get the typings for the exported function?

As a workaround, you can create and export a type for the function like so:

export type MyFunctionType = (param1: string) => boolean
export default (param1: string) => { return true; }

And then import like so:

import myFunction, { MyFunctionType } from './myFunction';

function bla(myFunction: MyFunctionType) {
  myFunction('hello')
}

but then you'd be specifying the type information twice, which is something I would like to avoid...

Answer 1

Louy was to some extend right. You need to use typeof to get the type information:

import myFunction from './myFunction'

function bla(myFunction: typeof myFunction) {
  myFunction('hello')
}

Answer 2

I think you're mixing types and default parameters or something.

Type of a is Function . a is not a type by itself. The default param can be specified like this:

function bla(aFn = a) {
  aFn('hello');
}

If you instead need aFn to have a signature similar to a , you'll have to specifiy that signature or create an interface for it.

function bla(aFn: (s: string) => boolean) {
  aFn('hello');
}

Or...

interface a { (s: string): boolean; }
function bla(aFn: a) {
  aFn('hello');
}