type TypeOfSomething = "something" | "other thing";
interface HigherLevel<T extends TypeOfSomething> {
subsection: MidLevel<T>;
}
interface MidLevel<T extends TypeOfSomething> {
callback: (arg: T) => void;
}
const t = { subsection: { callback: (arg: "something") => {} } } as HigherLevel;
This throws an error (typescript error) Generic type 'HigherLevel' requires 1 type argument(s).ts(2314)
Full Error: Generic type 'HigherLevel' requires 1 type argument(s).ts(2314)
On the line with as HigherLevel
Why is this? I have specified the type of arg
in the callback function so shouldn't typescript be able to infer the type parameter to HigherLevel
?
I am using as clause, I get a similar error when using it to type the constant t:
const t: HigherLevel = { subsection: { callback: (arg: "something") => {} } };
It gives me the same error
It would be nice for me to be able to have my types inferred, especially since I am only asking this question as it affects much more complicated typing I am trying to pull off.
I don't know if this is a feature of typescript, but it should be in my opinion or I might be thinking of this problem the wrong way.
Here is a simpler example without the HigherLevel
interface, which might be easier to solve. I am looking for a solution to the larger problem above still, but this still confuses me:
type TypeOfSomething = "something" | "other thing";
interface MidLevel<T extends TypeOfSomething> {
callback: (arg: T) => void;
}
const t: MidLevel = { callback: (arg: "something") => {} };
Same error, same line :(
TS version: 4.7.4 tsc -V
I generally use functions to infer generic types
function inferType<V extends TypeOfSomething>(t: HigherLevel<V>) { return t }
const t = inferType({ subsection: { callback: (arg: "something") => {} } })
Doubt there's another method