Comments (3)
@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.
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from typescript.