RequiredProps<T>: Unexpected result when T ={}

[yortus]

TypeScript Version: 2.8.0-dev.20180216

Code

// Type-level filters to extract just the required or optional properties of a type
// Defns from https://github.com/Microsoft/TypeScript/pull/21919#issuecomment-365491689
type RequiredPropNames<T> = { [P in keyof T]: undefined extends T[P] ? never : P }[keyof T];
type OptionalPropNames<T> = { [P in keyof T]: undefined extends T[P] ? P : never }[keyof T];
type RequiredProps<T> = { [P in RequiredPropNames<T>]: T[P] };
type OptionalProps<T> = { [P in OptionalPropNames<T>]: T[P] };

// Some object types with different numbers of props
type P2 = {a: string, b: number}; // Two props
type P1 = {a: string}; // One prop
type P0 = {}; // No props

// Let's extract only the required properties of P0, P1, and P2
type P2Names = RequiredPropNames<P2>; // P2Names = "a" | "b" ✓😊
type P1Names = RequiredPropNames<P1>; // P1Names = "a" ✓😊
type P0Names = RequiredPropNames<P0>; // P0Names = any ?😕

type P2Props = RequiredProps<P2>; // P2Props ={a: string; b: number} ✓😊
type P1Props = RequiredProps<P1>; // P1Props ={a: string} ✓😊
type P0Props = RequiredProps<P0>; // P0Props ={[x: string]: any } ?😕

Expected behavior:
P0Names = never and P0Props ={}

Actual behavior:
P0Names = any and P0Props ={[x: string]: any}

Notes:
@ahejlsberg helpfully came up with the RequiredProps and OptionalProps definitions above from a question I asked in #21919. They work great except when T ={}

Not 100% sure if this is a bug or not, but it's definitely counterintuitive. RequiredProps works for any object type except the empty one, when a string indexer suddenly appears. It certainly breaks the semantic expectations of RequiredProps.

Workaround:
The following version of RequiredPropNames achieves the expected behaviour by singling out the {} case:

type RequiredPropNames<T> =
 keyof T extends never
 ? never
 : {[P in keyof T]: undefined extends T[P] ? never : P}[keyof T];

[falsandtru]

This means {}[never] should return never.

[yortus]

@falsandtru I agree. Actually I think T[never] should return never for any type T.

Index access notation T[K] is valid for all K extends keyof T. Given that never extends keyof T for all possible types T, then T[never] should always be a valid index access type. Its result should always be never, since the union of zero keys (never) maps to the union of zero values (never).

[yortus]

In #11929 (the PR introducing indexed access types) it states:

An indexed access type T[K] requires K to be a type that is assignable to keyof T

...and in #8652 (the 'never' PR) it states that "never is a subtype of and assignable to every type."

If K is a union type K1 | K2 | ... | Kn, T[K] is equivalent to T[K1] | T[K2] | ... | T[Kn]

...and if n=0, then we effectively have an empty union / union of zero types, which is the type having zero values, which is never.

---

Well that's my reasoning anyway. #11929 doesn't mention never, but I believe the reasoning here is consistent with the given type rules.

[weswigham]

cc @sandersn didn't we recently change how T[any] behaved? Did we just leave out T[never]?

[yortus]

Closing in favour of #22042, which more simply addresses just the underlying issue.

[sandersn]

@weswigham I removed the constraint of T[string], which used to be any for unconstrained T. Maybe we have some more recent changes than that though?