21 Matching Annotations
  1. Feb 2023
    1. If you want a workaround for the case where you can't just replace key with a string literal, you could write your own user-defined type guard function called hasProp(obj, prop). The implementation would just return prop in obj, but its type signature explicitly says that a true result should cause obj to be narrowed to just those union members with a key of type prop: function hasProp<T extends object, K extends PropertyKey>( obj: T, prop: K ): obj is Extract<T, { [P in K]?: any }> { return prop in obj; } and then in your function, replace key in a with hasProp(a, key): function f3(a: A) { const key = 'b'; if (hasProp(a, key)) { return a[key]; // okay } return 42; }
  2. Aug 2021
    1. function strictIsDog<T extends Dog extends T ? unknown : never>( // like <T super Dog> candidate: Dog | T // if Dog extends T then Dog | T is T ): candidate is Dog { // compiler recognizes that Dog | T can narrow to T return "bark" in candidate; } if (strictIsDog(animal)) {} // okay if (strictIsDog(dog)) {} // okay if (strictIsDog(mixed)) {} // okay if (strictIsDog(cat)) {} // error! // ~~~ <-- Cat is not assignable to Dog
    1. Regarding the error message, the predicate type must be assignable to the value type because the type guard is used to check whether a value with a less-specific type is in fact a value with a more-specific type. For example, consider this guard: function isApe(value: Animal): value is Ape { return /* ... */ } Ape is assignable to Animal, but not vice versa.
  3. Oct 2019
    1. These typeof type guards are recognized in two different forms: typeof v === "typename" and typeof v !== "typename", where "typename" must be "number", "string", "boolean", or "symbol". While TypeScript won’t stop you from comparing to other strings, the language won’t recognize those expressions as type guards.