Extend type predicate inference to work with certain type variables

🔍 Search Terms

type predicate inference generic

✅ Viability Checklist

• This wouldn't be a breaking change in existing TypeScript/JavaScript code
• This wouldn't change the runtime behavior of existing JavaScript code
• This could be implemented without emitting different JS based on the types of the expressions
• This isn't a runtime feature (e.g. library functionality, non-ECMAScript syntax with JavaScript output, new syntax sugar for JS, etc.)
• This isn't a request to add a new utility type: https://github.com/microsoft/TypeScript/wiki/No-New-Utility-Types
• This feature would agree with the rest of our Design Goals: https://github.com/Microsoft/TypeScript/wiki/TypeScript-Design-Goals

⭐ Suggestion

When a function would normally serve as a type predicate for a discriminated union, but the exact cases being selected by the type predicate are dictated by a type variable, it'd be nice if the type predicate could still be inferred.

📃 Motivating Example

type Thing = A | B | C
type A = { kind: 'A', id: string, /* ... */ }
type B = { kind: 'B', id: string, /* ... */ }
type C = { kind: 'C', id: string, /* ... */ }

function filterByKind<T extends Thing['kind']>(things: Thing[], kind: T) {
  return things.filter((it) => it.kind === kind);
}

// The below would ideally be typed as `A[]`, as the function passed to 
// `filter` would have an inferred type predicate of `it is Thing & { kind: T }`
const x = filterByKind([], 'A');

const someKind = Math.random() < .5 ? 'A' : 'B';
const y = filterByKind([], someKind); // (A | B)[]

💻 Use Cases

I think this comes up somewhat rarely, and the workaround is to simply manually write out the generic type predicate, but, at my company, we did find cases like this when we looked at our type predicates as part of upgrading to TS 5.5

Im Issue werden keine Implementierungsdateien, Tests oder Einstiegspunkte genannt. Beginne damit, die motivierenden Beispiele zu reproduzieren, und untersuche das bestehende Verhalten bei der Inferenz von Typprädikaten. Als erledigt gilt die Aufgabe, wenn die generischen filterByKind-Beispiele für 'A' A[] und für eine Vereinigung von Arten (A | B)[] inferieren, ohne die Laufzeitausgabe zu ändern.