CODEWITHSUNDEEP
typescript
Juraj Garaj
Juraj Garaj

Reputation: 113

How to declare type for a function, that returns function with inferred type in typescript?

I'm trying to create type for functions that look like this:

const someSelector1 = (state: ContextState) => () => state.currentName; 

const someSelector2 = (state: ContextState) => (id: string) => state[id].values;

I would like to create type that would be assigned to the "someSelector" functions, and that would infer types of the return function. Something like this:

type Selector = <F extends Function>(state: ContextState) => F

const someSelector1: Selector = (state) => () => state.currentName; 

const someSelector2: Selector = (state) => (id: string) => state[id].values;

Unfortunately, with this implementation I'm getting Type '(id: string) => string' is not assignable to type 'F'. '(id: string) => string' is assignable to the constraint of type 'F', but 'F' could be instantiated with a different subtype of constraint 'Function'. error.

Is there a correct way to do this?

Upvotes: 1

Views: 65

Answers (1)

Robert Moore
Robert Moore

Reputation: 2572

You can't make a Selector type with no generics, but consider where you take in your Selector-typed arguments. In the function that uses the selectors, you can annotate the return type to give the return type of your selector like this:

type Selector<T extends () => any> = (state: ContextState) => T;

function applySelector<T extends () => any>(selector: Selector<T>): ReturnType<T> {
  // code
}

Now calling applySelector with a selector that returns T will also return T.

Upvotes: 1

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