typescript function generic parameter in spread argument

Question

I have the following function:

type SomeType<T> = Record<string, T>;
declare function test<T>(...args: SomeType<T>[]): T;

I would like to call it like that (a should be of type number|string):

const a = test({a: 3}, {b: "as"});

but that way I get an error: Type 'string' is not assignable to type 'number'. can I make this work without explicitly passing the generic argument number|string or using conditional types?

Playground

I'm fairly sure you're stuck with writing const a = test<string|number>({a: 3}, {b: "as"});, but not sure enough to make it an answer. :-) — T.J. Crowder
– T.J. Crowder, Commented Jun 5, 2021 at 8:36
oh yeah sorry, i forgot that in my question i dont want to explicitly add the generic on every call. Why cant typescript not just infer it as a union type. I mean its possible as i suggested using conditional types (as i have shown in the playground, but thats also really messy) — Hexception
– Hexception, Commented Jun 5, 2021 at 8:38
Don't worry, your question pretty strongly implied you didn't want to do that. :-) It's just I think you're stuck with it. TypeScript only goes so far with inference. — T.J. Crowder
– T.J. Crowder, Commented Jun 5, 2021 at 8:40

Roberto Zvjerković · Accepted Answer · 2021-06-05 09:58:45Z

1

Variadic tuples might help:

declare function test<T extends any[]>(...args: [...T]): T;

const a = test({ a: 3 }, { b: "as" });

answered Jun 5, 2021 at 9:58

Roberto Zvjerković

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2 Comments

Kas Elvirov Over a year ago

Just because it works, i wouldn't pass this solution to any production code because any, really means any typescriptlang.org/docs/handbook/2/everyday-types.html#any

Roberto Zvjerković Over a year ago

@ArtemVadimovichSolovev That was just to show how it could be accomplished, of course you might write extends Record<whatever> suits their need

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