Typescript: type guard for whether an object property is defined, when the key is a wide type?

Question

I have a function that returns whether an object property is undefined . I need this function instead of just doing obj[key] === undefined because otherwise I'd sometimes get Property 'foo' does not exist on type 'Bar'. . It's straightforward to write the type when the property key is a literal. Ie:

function hasDefinedProp<
  Obj extends Partial<Record<string, any>>,
  Prop extends string,
>(
  obj: Obj,
  prop: Prop,
): obj is Obj & Record<Prop, Prop extends keyof Obj ? Exclude<Obj[Prop], undefined> : unknown> {
  return obj[prop] !== undefined;
}

const obj: Partial<Record<string, number>> = {};
if (hasDefinedProp(obj, 'foo')) {
    obj.foo + 1; // No error.
    obj.bar + 1; // "Object is possibly 'undefined'."
}

However, this doesn't work when the key's type is a wide type, ie:

const obj: Partial<Record<string, number>> = {};
const key: string = '';
if (hasDefinedProp(obj, key)) {
    obj[key] + 1; // No error.
    obj.bar + 1; // No error. Should be "Object is possibly 'undefined'."
}

Is it possible to make the type guard work for wide types?

Answer 1

AFAIK, it is not possible. Once you have added explicit string<\/code> type to const key: string = '';<\/code> - TS is unable to narrow literal type of key<\/code> . As a result, you are allowed to use any string you want to access a property. TS is unable to distinguish two types with type string<\/code> in this example:

const obj: Partial<Record<string, number>> = {};
const key: string = '';
if (hasDefinedProp(obj, key)) {
    obj[key] + 1; // No error.
    obj.bar + 1; // No error. Should be "Object is possibly 'undefined'."
}