Optional properties on function interface breaks inference of type parameters

Bug Report

🔎 Search Terms

function interface optional properties generics type parameters inference

🕗 Version & Regression Information

This is the behavior in every version I tried, and I reviewed the FAQ for entries about type inference.

⏯ Playground Link

Playground link with relevant code

💻 Code

interface FuncA<T> {
  (arg: T): void;
};

interface FuncB<T> {
  (arg: T): void;
  x?: string;
};

const outerA = <T,>(func: FuncA<T>, arg: T) => {};
const outerB = <T,>(func: FuncB<T>, arg: T) => {};
const inner = <T extends 'a' | 'b' | 'c'>(cb: (arg: T) => void) => {};

outerA(inner, (arg: 'a' | 'b') => {});
outerB(inner, (arg: 'a' | 'b') => {}); // error

🙁 Actual behavior

In the call to outerA(), the type of the func parameter is inferred as FuncA<(arg: 'a' | 'b') => void>, as determined by the type of the arg parameter.

In the call to outerB(), the type of the func parameter is inferred instead as FuncB<(arg: 'a' | 'b' | 'c') => void>, using the base constraint of inner's T parameter rather than the subtype of that constraint used by the arg parameter.

🙂 Expected behavior

I do not expect the addition of an optional property on FuncB<T> to break the type parameter inference. It seems that inner's T parameter should be inferred as 'a' | 'b' in both cases.

Commencez par reproduire le comportement dans le TypeScript Playground lié en utilisant les exemples FuncA, FuncB, outerA, outerB et inner. Étudiez l’inférence des paramètres de type pour les deux appels et vérifiez la complétion lorsque, dans les deux cas, T de inner est inféré comme 'a' | 'b', sans l’erreur affichée pour outerB.