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(Not an expert in TS, this is just my understanding. Take it with a grain of salt). `const b = aOrB.kind ==="b" && b` assigns a new variable with a known type of `B`. `const isB = b.kind === "b"` assigns a variable of type `boolean` not `b is B`.
Since it's a bool, not a type guard, `isB &&` doesn't have the effect of changing the type of `aOrB`. As far as I can tell there's no way to assign a type guard to a variable.
I think the key is that the type of `b` is `B | undefined`, which is a simple type that doesn't depend on any of this type guarding business. So when the compiler sees `b && ...`, it doesn't know or care *why* `b` may be undefined.
You mean the type of `definitelyB`, right? It’s false | B, but same effect as if it were undefined | B
I wonder how f there's some way to preserve the type guard return type? I tried to use identity function <T>(a: T): T => a or ReturnType<typeof Guard> , to no avail
Opening an issue?
ohhhhhh.... I understand. I was confused why you put what looked like an existence check after checking aOrB.thing - very nicely done
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