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Comments (4)

Andarist avatar Andarist commented on April 27, 2024 1

Conditional types are tricky as they often stay deferred in generic contexts. This version could somewhat easily work though (and it doesn't):

type StringKeys<T> = keyof {
  [P in keyof T as T[P] extends string ? P & string : never]: any;
};

const bloop = <T,>(z: StringKeys<T>) => {
  const f: string = z; // errors but shouldnt
};

As a workaround, you can use this version:

type StringKeys<T> = keyof {
  [P in keyof T as T[P] extends string ? P : never]: any;
} &
  string;

const bloop = <T,>(z: StringKeys<T>) => {
  const f: string = z;
};

from typescript.

fatcerberus avatar fatcerberus commented on April 27, 2024 1

I would guess it’s just a matter of the compiler knowing that T & string is always assignable to string regardless of what T is (by definition, as A & B is the set of values assignable to both)

from typescript.

mikeauclair avatar mikeauclair commented on April 27, 2024

That workaround works in my real codebase, so I'll cut over to that approach for now - is this a matter of the intersection (instead of Extract) causing non-deferred evaluation?

from typescript.

RyanCavanaugh avatar RyanCavanaugh commented on April 27, 2024

Bisects to #56742

from typescript.

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