Result type for JSON parsing in TypeScript
TypeScript interview exercise on TypeScript, difficulty 3 of 3. Solve it in the editor beside a worked example; the tests run the moment you do.
Declare the same Result<T> type, then write safeParse<T = unknown>(json: string): Result<T> that never throws: on success return { ok: true, value }, on failure { ok: false, error } with the parser's error message.
Each exercise pairs a worked example with a twin problem to solve in a real editor; the tests run the moment you do. Progress is saved in your browser, and in your account if you sign in.