BC break for method call on union type

[Strate]

TypeScript Version: 2.1.0-dev.20160729

Code

interface A {
 focus: (force: boolean) => void
}
interface B {
 focus: () => void
}
let instance: A | B
instance.focus(true);

Expected behavior:
No compilation error at least at 2.1.0-dev.20160716

Actual behavior:
TS2349: Cannot invoke an expression whose type lacks a call signature.

[Strate]

Maybe similar case, also broken between 2.1.0-dev.20160716 and 2.1.0-dev.20160729:

interface List<T> {
 map<U>(mapper: (v: T, index: number) => U): List<U>
}
interface A {
 prop1: boolean
}
interface B extends A {
 prop2: number
}
type ListA = List<A>;
type ListB = List<B>;
let instance: List<A> | List<B>;
instance.map((v, index) => {
 return true;
})

[RyanCavanaugh]

Anders Hejlsberg (@ahejlsberg) I believe this is due to the less-aggressive subtype reduction -- thoughts?

[ahejlsberg]

Artur Eshenbrener (@Strate) I'm curious what the scenario is for having types like List<A> | List<B> when B is a subtype of A. You could equally well just write List<A> (and the issue would then go away). Similarly, you could just use type A instead of A | B in the first example.

Ryan Cavanaugh (@RyanCavanaugh) Yes, this is caused by a change in #9407 to not perform subtype reduction in property accesses on objects of union types. Subtype reduction is horribly expensive on large union types and it is not clear we could ever make it perform acceptably. Also, subtype reduction only helps in those cases where we can reduce a union of function types down to a single function type.

The core issue really is that we are very conservative when it comes to call operations on a union type. We basically require all of the parameter types to be identical, with a small allowance matching omitted parameters to optional parameters.

The real fix here would be to find a better way to check calls on unions of function types. Subtype reduction isn't the solution because there are lots of cases where we have a union of function types that aren't subtypes of each other, but where there is a meaningful way to construct calls that would satisfy all of the signatures.

One possibility might be to intersect the types of parameters in corresponding positions. So, for a union of function signatures (() => T) | ((x: A) => U) | ((x: B, y: C) => V) we would produce a synthetic signature of type ((x: A & B, y: C) => T | U | V).

[Strate]

Anders Hejlsberg (@ahejlsberg) seems that your suggestion could solve this issue:

type a = string;
type b = number;
interface List<T> {
 map<U>(mapper: (item: T) => U): List<U>
}
let instance: List<a> | List<b>
instance.map(v => true)

call to map produces an error "TS2349: Cannot invoke an expression whose type lacks a call signature." and "TS7006: Parameter 'v' implicitly has an 'any' type.". Second one could be fixed by explicit add a|b type for v parameter, but first one fixable only by typecasting instance to List<a|b>

[RyanCavanaugh]

#10041 is a good example of how this can be problematic in the absence of a meaningless union type

[ahejlsberg]

Artur Eshenbrener (@Strate) Yes, but as I think about it more, the idea of intersecting the parameter types still wouldn't work for overloaded signatures (as it wouldn't be clear which signatures to match with other signatures).

Still curious whether this shows up in a real world scenario for you, and if so, what leads to the existence of union types where one constituent is a subtype of the other?

[Strate]

It seems that there is no difference between

interface A { prop1: boolean }
interface B extends A { prop2: boolean }

and

interface A { prop1: boolean }
interface B { prop1: boolean, prop2: boolean }

from structural type system view.
First case is just to reduce copy-paste props from A to B.
And there is definetely big difference between Array<A> | Array<B> and Array<A|B> and, of course, Array<A>.

What about real-world scenario, I can't formalize it now, maybe our sode really could be refactored to your suggestion :)

[zpdDG4gta8XKpMCd]

I had a similar break in my code:

type Rule = LocalRule | GlobalRule;
interface LocalRule { enforce: () => void; }
interface GlobalRule { enforce: (state: GlobalState) => void; }
declare rules: Rule [];
rules.forEach (rule => rule.enforce (state)); // used to work, not anymore

[zpdDG4gta8XKpMCd]

To be frank, the only reason I wrote it this way is because of my laziness, I didn't want to pattern match to a certain case as long as the unified signature would work as well.

[yortus]

Aleksey-Bykov laziness was smart in this case since pattern matching with LocalRule|GlobalRule also has a subtype reduction problem:

rules.forEach(rule => {
 if (isGlobalRule(rule)) {
 rule.enforce(state);
 }
 else {
 rule.enforce(); // ERROR: Property 'enforce' does not exist on type 'never'.
 }
});

[zpdDG4gta8XKpMCd]

Troy Gerwien (@yortus) the way i fixed is by adding a kind property to both cases that i used for discrimination, without such property i see how it can be a problem, good catch