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Refining discriminated union of arrays with array.find on a property that is a union #60785
Description
🔎 Search Terms
"discriminated union", "refine union", "array find union property"
🕗 Version & Regression Information
Tried every TypeScript version from 5.2.2 in the playground with the same result.
I found some similar issues that eventually led me to #42384, but I wasn't sure if the root cause is the same?
⏯ Playground Link
💻 Code
interface Sharp {
// Removing 'Triangle' here to get rid of the union makes it work.
type: 'Square' | 'Triangle';
corners: number;
}
interface Round {
type: 'Circle' | 'Oval';
radius: number;
}
function getCorners(shapes: Sharp[] | Round[]) {
// Expected `Sharp | undefined` but got `Sharp | Round | undefined`
const foundSharp = shapes.find((shape) => shape.type === 'Square');
// Property 'corners' does not exist on type 'Sharp | Round'.
// Property 'corners' does not exist on type 'Round'. (2339)
return foundSharp?.corners;
}🙁 Actual behavior
foundSharp is Sharp | Round | undefined even though it can never be Round due to type.
🙂 Expected behavior
foundSharp should be Sharp | undefined since only sharp shapes can have the type 'Square'.
Additional information about the issue
Setting type in the Sharp interface to only 'Square' makes it work as expected.
Activity
jcalz commentedon Dec 17, 2024
This isn't a TS bug. Your callback isn't inferred as returning a type predicate as implemented in #57465, because it would need to be a one-sided type predicate, which TypeScript doesn't have. The check
shape.type === 'Square'can narrowRound | SharptoSharpbut sinceshape.type !== 'Square'does not narrowRound | SharptoRound, no type predicate can be inferred. (SeeflakyIsStringin the text of #57465) If TS had one-sided type predicates as requested in #15048, then the return type could be inferred as something likeshape is Shape else falseand it would work. When you're usingfind()you don't care about the false side of the type guard function, but TS does, and there's currently no way to tell it not to.If you want, you can annotate your callback with a type predicate return type, and while
shape is Sharpwill work, it's a bit of a lie. Something which works and is true is:Playground link to code
elusiveunit commentedon Dec 18, 2024
I see, thanks for the explanation! We'll make do with some helpers then and keep an eye on the one-sided predicate issue.