Return type inference of overloaded function when argument is known

[KSXGitHub]

TypeScript Version: 3.0.3

Search Terms:

Code

declare function fn(x: 'a'): 0
declare function fn(x: 'b'): 1
declare function fn(x: 'c'): 2
declare function fn(x: 'D'): 3

// expected: type One = 1
// received: type One = never
type One = (typeof fn) extends (x: 'b') => infer Return ? Return : never
const one: One = 1

Expected behavior:

One should be 1

Actual behavior:

One is never

Playground Link: link

Related Issues: #26591

[RyanCavanaugh]

Overloads aren't resolved during inference (only the last is examined), nor are they expanded out combinatorially. In general for any given sequence of specific overloads like this there "should" be a final signature like (x: 'a' | 'b' | 'c' | 'D'): 0 | 1 | 2 | 3, which will be the one used during inference. This won't give you the most specific type, but would prevent hitting never here

Example without conditional types:

function call<T>(f: () => T): T {
	return f();
}

declare function f(): string;
declare function f(x: string): void;
const x = call(f); // Want string, got void

[huan]

The "Example without conditional types" posted by @Andy-MS is still true under TypeScript v4.4.4

I believe this is definitely typing system bug.