Typescript ReturnType of generic function

Question

The new ReturnType in TypeScript 2.8 is a really useful feature that lets you extract the return type of a particular function.

function foo(e:          


        

Answer 1

TypeScript compiler does not see typeof foo as generic type. I'd say it's a bug in the compiler.

However, TypeScript has callable interfaces which can be generic without any problems, so if you introduce a callable interface compatible with the signature of your function, you can implement your own equivalent of ReturnType like this:

function foo<T>(x: T): T {
  return x;
}


interface Callable<R> {
  (...args: any[]): R;
}

type GenericReturnType<R, X> = X extends Callable<R> ? R : never;

type N = GenericReturnType<number, typeof foo>; // number

Answer 2

If you want to get some special generic type, You can use a fake function to wrap it.

const wrapperFoo = () => foo<number>()
type Return = ReturnType<typeof wrapperFoo>

More complex demo

function getList<T>(): {
  list: T[],
  add: (v: T) => void,
  remove: (v: T) => void,
  // ...blahblah
}
const wrapperGetList = () => getList<number>()
type List = ReturnType<typeof wrapperGetList>
// List = {list: number[], add: (v: number) => void, remove: (v: number) => void, ...blahblah}

Answer 3

This is my currently working solution for extracting un-exported internal types of imported libraries (like knex):

// foo is an imported function that I have no control over
function foo<T>(e: T): InternalType<T> {
    return e;
}

interface Wrapper<T> {
  // wrapped has no explicit return type so we can infer it
  wrapped(e: T) {
    return foo<T>(e)
  }
}

type FooInternalType<T> = ReturnType<Wrapper<T>['wrapped']>
type Y = FooInternalType<number>
// Y === InternalType<number>

Answer 4

I found a good and easy way to achieve this if you can change the function definition. In my case, I needed to use the typescript type Parameters with a generic function, precisely I was trying Parameters<typeof foo<T>> and effectively it doesn't work. So the best way to achieve this is changing the function definition by an interface function definition, this also will work with the typescript type ReturnType.

Here an example following the case described by the OP:

function foo<T>(e: T): T {
   return e;
}

type fooReturn = ReturnType<typeof foo<number>>; // Damn! it throws error

// BUT if you try defining your function as an interface like this:

interface foo<T>{
   (e: T): T
}

type fooReturn = ReturnType<foo<number>> //it's number, It works!!!
type fooParams = Parameters<foo<string>> //it also works!! it is [string]

//and you can use the interface in this way
const myfoo: foo<number> = (asd: number) => {
    return asd;
};

myfoo(7);