How to make TypeScript infer number & types of arguments from an array

Question

I try to create a dead-simple function that accepts:

A callback
An optional array of arguments to pass to the callback

The function in plain JS would look like this:

const callCallback = (cb, args = []) => cb(...args);

The callback might accept an arbitrary number of arguments & return whatever. The problem is that I can't properly express the type of callback with TypeScript. I tried to declare callCallback's type like this (the idea was to infer callback's arguments & return types):

export interface CallCallback {
  <CbReturn, CbArgs>(
    cb: (...optionalArguments: CbArgs[]),
    args?: CbArgs[],
  ): CbReturn
};

However, this didn't work because it forces the callback to use arguments spreading syntax while it's not what I intend to do.

I also found this answer but as for now TS doesn't let me make spread argument optional.

How can I infer args number & types into cb parameters list?

y2bd · Accepted Answer · 2021-01-30 01:52:15Z

3

Would the below work? It seems like you're essentially typing apply:

interface ApplyFn {
    <F extends () => any>(fn: F): ReturnType<F>;
    <F extends (...args: any[]) => any>(fn: F, args: Parameters<F>): ReturnType<F>;
}

declare const apply: ApplyFn;

apply((f: number, s: string, t: Date) => f + s + t, [23, "two", new Date]); // OK
apply((f: number, s: string, t: Date) => f + s + t, ["not a number", "two", new Date]); // ERR
apply((f: number, s: string, t: Date) => f + s + t, [123, "two"]); // ERR

apply(() => "nice!", []); // OK
apply(() => "nice!"); // OK
apply((param: number) => "nice!"); // ERR

Edit: The type above is useful if you have a hole for this type of function to fit in (e.g., a function argument). If you're actually trying to define this function, it's easier to represent it as a standard function definition:

function apply<F extends () => any>(fn: F): ReturnType<F>;
function apply<F extends (...args: any[]) => any>(fn: F, args: Parameters<F>): ReturnType<F>;
function apply(fn: Function, args: any[] = []) {
  return fn(args);
}

edited Jan 30, 2021 at 1:52

answered Jan 29, 2021 at 20:31

y2bd

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4 Comments

Igor Bykov Over a year ago

Thank you, so much! I don't completely get how it works because it's a way too advanced for me, but your answer certainly helped a lot.

y2bd Over a year ago

Feel free to point out what you don’t understand and I’ll expand my answer.

Igor Bykov Over a year ago

That would be great, first of all, could you please add an actual function declaration in your answer? Because declare const callCallback: ApplyFn = (fn, args = []) => fn(...args); produces a type error & I can't figure out how to apply ApplyFn type.

y2bd Over a year ago

I added the most simple way to write the actual function in TS.

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