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Partially overlapping union discriminators are not assignable from a union of discriminating values #45230
Description
Activity
between v3.3.3 and v3.5.1 something changed
The change was "smarter union type checking", implemented in #30779 and released in TS3.5.
In general if you have some type
{ a: U1, b: U2, c: U3, d, U4 }, you can't tell if that's going to be assignable to some target type without doing a computation involvingsize(U1) * size(U2) * size(U3) * ...different expansions of the type to enumerate all possible inhabitants. So we're always going to sometimes reject some of these assignments because the special-casing to allow the trivial cases has some limits.That said, I don't see why this should matter when a non-literal type appears in the target.
This is the expected behavior. It is far too expensive to compute this kind of overlapping assignability for all cases, so we only currently do this work when a union has overlapping discriminant properties. A property is a discriminant property when its type consists of only unit types (e.g.,
1or"foo"). In the three failing cases above, the property in question consists of a unit type unioned with a non-unit type (string,RegExp, etc.), which means the property is not considered a discriminant and thus does not qualify for this assignability rule.In the case of
C, both types in the union contain discriminant properties as they consist only of unit types or unions of unit types (1 | 3and2 | 3).
Bug Report
🔎 Search Terms
tagged, discriminated, union, partially, overlapping, infinite, types, not, assignable
🕗 Version & Regression Information
I tried versions from v3.3.3 to v4.3.5, as those were available in the Playground at this time.
Note that between v3.3.3 and v3.5.1 something changed that affects type
Cfrom my example below. In v3.3.3 the assignment toconst c: Cis an error, from v3.5.1 its okay.⏯ Playground Link
Playground link with relevant code
💻 Code
🙁 Actual behavior
The assignments to
const a: A,const b: B, andconst d3: Draise an error.🙂 Expected behavior
As demonstrated in the assignments to
const d1: Dandconst d2: D, my partially discriminating propertydpropcan be either of the two discriminating values. So It should be possible to assign todpropa value that is either of them, as in the assignment toconst d3: D.