Allow a type guard function to also return a boolean literal whenever possible about typescript HOT 3 CLOSED

@whzx5byb wow, that is so much closer than I have been able to get. I'm inclined to say that that is good enough to be able to close this issue.

I have been trying to see if I can create a second overload to cover the false case, but I'm not managing that one just yet. Although that's now a little less important, because I can just infer myself that if the function's return-type is boolean instead of true, that I should interpret that as false.

But, if you happen to be willing to try to help me achieve the false case as well, then I'd be super grateful!

Edit

@whzx5byb never mind! I figured it out! Here it is!

from typescript.

I think I've answered all of these questions above, didn't I?

You didn't answer the most important one: what is the use case? AFAICT, in any case where you can know at compile time what the result of a type guard is going to be, then there's no reason to induce runtime overhead by calling it.

from typescript.

Does Function Overloading work for you?

type Primitive = string | number | bigint | boolean | symbol | null | undefined
declare function isPrimitive<T extends Primitive>(thing: T): true;
declare function isPrimitive(thing: unknown): thing is Primitive;

// Example usage:
const value: unknown = "hello";
if (isPrimitive(value)) {
  value // value is narrowed to be a `Primitive`
  // ^? const value: Primitive
}

const r = isPrimitive(123) // returns the boolean literal `true`
//    ^? const r: true

from typescript.