When working with typescript, the following code will be typed correctly:
let v: number | null | undefined;
if(v === null || v === undefined) return;
// v is now recognized as a `number`
const v2 = v + 2;
However, if we decide to streamline this process by creating a helper function for the null and undefined check, Typescript may raise an issue:
let v: number | null | undefined;
const isNull = (v: number | null | undefined) => v === null || v === undefined;
if(isNull(v)) return;
// In this case, v is typed as `number | null | undefined`
const v2 = v + 2;
Is there a way to pass along the equality check for null and undefined when using the helper function isNull?
One workaround could be to override it with v!, but this might seem unorthodox and risky as it creates decoupling between the two.