CODEWITHSUNDEEP
typescripttypescript-generics
michaeloliver
michaeloliver

Reputation: 764

How to infer type of function parameters in generic type

I am trying to create tuple type that contains two elements, the first being a function and the second the parameters it takes. I have been somewhat successful by creating a generic type CallbackTuple. Ideally I would like to be able to infer the second element of the tuple without explicitly having to pass the type argument typeof a, how would I do this?

function a({ foo, bar }: { foo: string; bar: number }) {
  return { foo, bar };
}

type CallbackTuple<
  PluginFn extends (...args: any[]) => any = (...args: any[]) => any
> = [PluginFn, ...Parameters<PluginFn>];

const cbTuple1: CallbackTuple<typeof a> = [a, { foo: "", bar: 1 }]; // Valid
const cbTuple2: CallbackTuple<typeof a> = [a, { foo: "" }]; // Property 'bar' is missing
const cbTuple3: CallbackTuple = [a, { foo: "" }]; // I want this to type check properly

Upvotes: 1

Views: 2501

Answers (1)

Titian Cernicova-Dragomir
Titian Cernicova-Dragomir

Reputation: 249466

You can't do this without an extra function do help with inference. Variables can either have their types specified or inferred. There is no way to have just some parts of a variable type inferred.


function makeCallbackTuple<T extends (...args: any[]) => any>(p: CallbackTuple<T>) {
    return p
}

const cbTuple1 = makeCallbackTuple([a, { foo: "", bar: 1 }]); // Valid
const cbTuple2 = makeCallbackTuple([a, { foo: "" }]); // Property 'bar' is missing

Playground Link

Upvotes: 1

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