How can I force typescript to only allow code to execute if two types are strictly equal to each other?

Question

typescript

Let's say we have code like this:

class UserModel {
    // Все те поля, что есть у UserDTO 
    password: string;
} // Модель User в БД. 

class UserDTO {
    email: string;
}

const getUser = async (email: string): Promise<UserDTO> => {
     const user = await db.find(email); // Тип возращаемого объекта является UserModel

      return user; // UserModel не эквивалентен UserDTO, но TypeScript не считает это ошибкой.
}

How to prevent typescript from allowing returning an object equivalent to this type as a UserDTO, except that this object also has additional fields that are not described in UserDTO.

Answer 1

Answer: don't use typescript.
Structural typing is the essence of typescript. You can google all sorts of vile crutches, but they are not a panacea and will fail at the most crucial moment.

Answer 2

The easiest way is to make the types unequal. For example like this:

class UserModel {
    email: string;
    password: string;
}

class UserDTO {
    protected __guard: never; // добавляем фиктивное поле
    email: string;
}

const getUser = async (email: string): Promise<UserDTO> => {
  const user = await db.find(email);
  return user; // Ошибка, как заказывали
}