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typescripttypescript-generics
Neo
Neo

Reputation: 4760

Typescript Generic Function Interface in another interface

I have a generic interface

export interface Method<T> {
    (answer: T): T;
}

I want to create an interface where I have fn function

interface Wrapper {
    fn: <T,>(a: T ) => T
}

but instead of manually writing the function type I want to use the interface

interface MethodWrapper {
   fn: Method
}

I want the fn to be generic at fn level not at the interface level. Is it possible?

ts-playground link

Upvotes: 2

Views: 696

Answers (2)

jcalz
jcalz

Reputation: 330456

TypeScript's type system lacks the expressiveness to programmatically transform Method<T> into Wrapper. You'd need to move the scope of the generic type parameter T around in ways that can't be done.

If TypeScript had generic values as requested in microsoft/TypeScript#17574, then you might be able to say something like:

// Invalid TypeScript, don't do this
interface Wrapper {
  fn: forall T. Method<T>;
}

or

// Invalid TypeScript, don't do this
interface Wrapper {
  fn: <T> Method<T>;
}

But it doesn't so we can't.

You can almost do it the other way around and define Method<T> in terms of Wrapper, using instantiation expressions. I say "almost" because it requires that you drop down to the type level and acquire (or pretend to acquire) a value wrapper of type Wrapper before lifting back up to the type level. Like this:

declare const wrapper: Wrapper; // pretend to have this

type Method<T> = typeof wrapper.fn<T> // instantiate fn with <T>
// type Method<T> = (a: T) => T

Depending on your use case that may or may not be helpful.

Playground link to code

Upvotes: 2

raz-ezra
raz-ezra

Reputation: 559

Not sure I understand your question fully, but guessing from your TS playground you need to pass the type of the generic:

interface MethodWrapper {
   fn: Method<T> // T should be the type you want fn to receive
}

Upvotes: 0

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