TypeScript cannot always infer the correct type through a user-defined type guard using the 'is' operator.

🔎 Search Terms

"typeguard", "is operator"

🕗 Version & Regression Information

Tested with typescript@5.1.3 and later

⏯ Playground Link

No response

💻 Code

enum kind {
	a = 'a',
	b = 'b',
	c = 'c'
}

type ID = {
	id: number
}

class Test<T extends kind> {
	value: T extends kind.a | kind.b ? ID : undefined
	type: T

	constructor(
		type: T,
		value: T extends kind.a | kind.b ? ID : undefined
	) {
		this.type = type
		this.value = value
	}

	hasID(): this is Test<Exclude<kind, kind.c>> {
		return this.type !== kind.c
	}

	do() {
		if (this.hasID()) {
			return this.value.id // Error: this.value: Object is possibly 'undefined'
			// return (this as Test<Exclude<kind, kind.c>>).value.id // this works
		}
	}
}

🙁 Actual behavior

TypeScript cannot infer that 'this' is of type 'Test<Exclude<kind, kind.c>>' even though there is a type guard. In the 'do' method, TypeScript throws a TSError stating that 'this.value': Object is possibly 'undefined'.

If we explicitly cast 'this' as 'Test<Exclude<kind, kind.c>>', we can access the 'value' attribute, and it works as it should. However, somehow the type guard does not have the same effect.

🙂 Expected behavior

Since there is a type guard around the 'this.value.id' line, TypeScript should handle 'this' as a 'Test<Exclude<kind, kind.c>>' type. At least, that's what type guards are for in my understanding.

Additional information about the issue

No response

Start with the supplied TypeScript reproduction and trace how the type checker narrows this after the hasID() user-defined type guard. Add a compiler regression test for the expected this.value.id access, then verify that the diagnostic is removed without weakening checks for the kind.c case.