type Check<S extends Record<unique, unknown>> = S;
type Output = Check<{
b: number;
}>;
By defining
S extends Record<unique, unknown>In Typescript, interfaces are considered to be "open", allowing for additional properties to exist on an object that are not defined within its interface. Surprisingly, this does not result in a compile-time error.
As a result, an object of type {a: string} can be assigned as a valid value for Record<symbol, unknown>.
To demonstrate this flexibility, you can use the following code snippet which will compile without any issues:
const exampleObj: {a: string} = {a: "bar"};
const recordExample: Record<symbol, unknown> = exampleObj;
To render it void, you can combine it with a different Record containing only values of never:
type Test<T extends Record<symbol, unknown> & Record<string | number, never>> = T;;
type Result = Test<{ // ! error, string/number keys should be type never
a: string;
}>;
Works pretty well
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