microsoft / TypeScript

TypeScript is a superset of JavaScript that compiles to clean JavaScript output.
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Reflect.has fails to act as type guard (should act same as "in" operator) #30688

Open lukewlms opened 5 years ago

lukewlms commented 5 years ago

TypeScript Version: ^3.4.0-dev.20190330

Search Terms: Reflect.has in operator

Code

const test1 = (a: { field: number } | {}) => (("field" in a) ? a.field : 0);

const test2 = (a: { field: number } | {}) => Reflect.has(a, "field") ? a.field : 0;

Expected behavior: Both compile successfully. "in" operator acts as a type guard per https://github.com/Microsoft/TypeScript/issues/10485. Reflect.has should act the same as "in" operator here.

We'd like to use Reflect.has a lot more and this case holds us back. Thank you!

Actual behavior: "in" operator line compiles; Reflect.has line does not compile. Error:

allPasos.ts:445:31 - error TS2339: Property 'field' does not exist on type '{} | { field: number; }'. Property 'field' does not exist on type '{}'.

445 Reflect.has(a, "field") ? a.field : 0;

Playground Link:

https://goo.gl/2R4dkS

Related Issues:

https://github.com/Microsoft/TypeScript/issues/10485

dragomirtitian commented 5 years ago

This could be a definition of has that would work as you expect:

declare namespace Reflect {
  function has<T, K extends PropertyKey>(target: T, propertyKey: K): target is Extract<T, Record<K, any>>;
}
const test2 = (a: { field: number } | {}) => Reflect.has(a, 'field') ? a.field : a;

As a workaround you could add it yourself.

lukewlms commented 5 years ago

Looks like that does work as a stopgap! Thanks for the suggestion, added that in our custom types code.