Bug with mapped types and indexing

[SimonMeskens]

Minimal repro:

type AB = {
 a: 'a'
 b: 'a'
}

type T1<K extends keyof AB> = { [key in AB[K]]: true }
type T2<K extends keyof AB> = T1<K>[K] // BUG: should be an error for K = 'b'

I came across this in production code and made a contrived example to show the issue. Might be related to #15756.

Note that there might be two bugs here in play, one with mapped types and one with lookup types on mapped types. Once I start pushing the types a little, I come across a wild range of unexpected behavior.

Pay attention to the last type 'G', where the compiler is suddenly correct again, which is probably the oddest one of all.

// Let's make a simple type

type StringA = "a"

type A = {
 a: StringA
 b: StringA
}

// so type A[S] should always resolve to "a" for any valid keys
// note that it doesn't collapse T, even though it could

type T<S extends "a" | "b"> = A[S] // expands A, but doesn't collapse the indexer

let t1: T<"a"> // t1: "a"
let t2: T<"b"> // t1: "a"

// Let's make a silly generic type that only keeps key "a"
// The compiler correctly infers that this new type is just{a: true }

type B<S extends "a" | "b"> = {[key in "a"]: true} // B ={a: true }

// Let's make the type more generic, remember that A[S] is actually just "a"
// The compiler now infers that the type of C is{}, this is plain wrong

type C<S extends "a" | "b"> = {[key in A[S]]: true} // C ={}

// Obviously, we couldn't do a lookup on such a type
// Compiler correctly complains that we can't index an empty object

type D<S extends "a" | "b"> = {}[S] // Type 'S' cannot be used to index type '{}'.

// Now let's add a lookup to the generic version
// Curiously, it doesn't fail this time, but infers 'true', how odd

type E<S extends "a" | "b"> = {[key in A[S]]: true}[S] // = true

// It even happens when we index the above type C, which the compiler claims is{}

type F<S extends "a" | "b"> = C<S>[S] // = true

// Let's also try to use this type, it will become important later
// This does exactly what's expected

let a1: F<"a"> // a: true
let b1: F<"b"> // b: true

// Now let's make it even weirder and add a general indexer
// This correctly doesn't fail as any key is now valid
// The type it infers though, is just plain weird:
// G = ({} &{[key: string]: false})[S]

type G<S extends "a" | "b"> = (
 C<S> &
 {[key: string]: false}
)[S]

// And now the weirdest thing of all, let's use this type
// It correctly infers the result! Both of these are correct!

let a2: G<"a"> // a: true
let b2: G<"b"> // b: false

This has to be the weirdest bug I've run across

[gcnew]

It looks like you are experiencing the effects of #12351. The strange behaviour is happening because some of the usecases are expanded, while others are handled on the logical level, without carrying out the heavy lifting.

[olegdunkan]

type C<S extends "a" | "b"> ={[key in A[S]]: true} - maybe in this line of code the compiler doesn't evaluate A[S] because of some of sort of lazy evaluation, why does it have to go deeper and analyze A[S] ? But when you type E<S extends "a" | "b"> ={[key in A[S]]: true}[S] you force compiler to find out exactly what are inside A<T>. It is my assumption.

[gcnew]

Because of #12351

The operation{[P in K]: T}[X] is equivalent to an instantiation of T where X is substituted for every occurrence of P. For example,{[P in K]: Box<T[P]>}[X] is equivalent to Box<T[X]>.

type E<S extends "a" | "b"> = {[key in A[S]]: true}[S]

is converted to just

type E<S extends "a" | "b"> = true

by applying the above rule, without doing any actual "mapping" or "selection".

PS: You can see the rule is in action, because "extra" keys are silently ignored:

type E<S extends "a" | "b" | "extra"> = {[key in A[S]]: true}[S] // no error

[gcnew]

As strange as it may seem

type C<S extends "a" | "b"> = {[key in A[S]]: true} // C ={}

is actually correct. The reason is that S extends "a" | "b" means that S can be either

• never
• "a"
• "b"
• "a" | "b"

TypeScript plays it safe here and chooses never. The "expected" type is returned when C is instantiated with a specific value for S.

declare const x: C<"a"|"b">; //{a: true } 

Note that the same applies for S extends keyof A.

[SimonMeskens]

I disagree that choosing never is ever the right choice. Every type has the option of becoming never. If we assume that, then every inferred type everywhere should become never.

Also, it should never collapse a parametrized type, as you can just wait until it's used and check what the correct value is. A type constraint is just that, a constraint. It should pick the widest type possible for constraints, so in this case, it should pick "a" | "b" as long as possible.

In any case, this behavior is clearly buggy and needs to be fixed.

[gcnew]

I disagree. The compiler should (and does) choose the most neutral type. In this case, because at use site the S parameter might be any of the above mentioned options, the only safe one is the least restrictive one - i.e. never. If another option was chosen, it would have forced every use site to provide the pertinent properties. If all properties is the intention the right operator is keyof. When talking about sets of options, the only choice that makes no assumptions is the empty set.

[SimonMeskens]

So how would you expect someone to write a parameter constrained by string keys? S extends keyof{a, b }?

Would that have different behavior from S extends "a" | "b"? Sounds like that would still cause buggy behavior, as keyof{a, b } infers to "a" | "b" when collapsed in other places.

[gcnew]

I'm not sure I understand you. The parameter is still constrained to the said set of keys, it's just that intelli-sense chooses one of the valid options which is the empty set. In subsequent uses of the alias, both value and type level, it works correctly. I'm not convinced that choosing the fullest set in intelli-sense would be better.

[SimonMeskens]

Ah, you're just talking about intellisense? I really don't agree. Intellisense should always display the widest possible type. Otherwise you run into cases where intellisense tells you a certain type is {} and then a line below it, suddenly it widens to a different type.

That doesn't change the fact that there's a clear bug going on here though, that needs to be fixed.

[gcnew]

Which one do you feel is a bug?

From the comments I'm left with the impression that

type C<S extends "a" | "b"> = {[key in A[S]]: true} // C ={}

and

type E<S extends "a" | "b"> = {[key in A[S]]: true}[S] // = true

are the ones you feel are odd, but I think I've explained them in #17238 (comment) and #17238 (comment).

Edit: I've updated wording and gave more context.

[SimonMeskens]

I think the example above provides ample amount of bugs, but the main one is this:

type B<S extends "a" | "b"> = {[key in "a"]: true} // B ={a: true }
type C<S extends "a" | "b"> = {[key in A[S]]: true}
type F<S extends "a" | "b"> = C<S>[S] // = true

let a1: F<"a"> // a: true
let b1: F<"b"> // b: true

How can b1 ever be true? That's just an outright bug. Type C is exactly the same as type B, because A[S] is always "a". So type C is {a: true }. b1 tries to index with an index that doesn't even exist on the type it's indexing.

[gcnew]

b1 tries to index with an index that doesn't even exist on the type it's indexing.

You use double indexing. b resolves to an "a".

type StringA = "a"

type A = {
 a: StringA
 b: StringA
}

[gcnew]

OK, I see. Yes, it is a bug. The problem is what happens here:

PS: You can see the rule is in action, because "extra" keys are silently ignored:

type E<S extends "a" | "b" | "extra"> = {[key in A[S]]: true}[S] // no error

A[S] should still be checked for compatibility before applying the "substitution" rules.

[SimonMeskens]

I'll add one more example to make it clear:

type F<S extends "a" | "b"> = C<S>[S] // = true

let a1: F<"a"> // a: true
let b1: F<"b"> // b: true

type G<S extends "a" | "b"> = (
 C<S> &
 {[key: string]: false}
)[S]

let a2: G<"a"> // a2: true
let b2: H<"b"> // b2: false

Adding a bogus indexer to F makes it give different results. I'll edit G above with this version to make it clearer.