When to Use a Private Constructor | TypeScript OOP
One of the first things we learn when we start out is how to create instances of objects. Typically, we do this with the new keyword.
class User {
public name: string;
constructor (name: string) {
this.name = name;
}
}
const user: User = new User('Khalil Stemmler');The actual thing that does the creating is the constructor- and by default, it's public.
Have you ever seen a private constructor? Changing the scope of a constructor to private removes our ability to use the new keyword.
class User {
public name: string;
private constructor (name: string) {
this.name = name;
}
}
const user: User = new User('Khalil Stemmler'); // ErrorWhy on Earth would you want to do that? We can't create instances anymore. How are we supposed to get Users out of this class now?
All is not lost, and it turns out that there is a very good reason for why you'd want to do this kind of thing.
In essence, it's to enforce object creation rules.
Using the new keyword
When we use the new keyword, there's not really an elegant way to prevent a User from being created if certain validation rules don't pass.
We can throw errors.
class User {
public name: string;
constructor (name: string) {
if (!!name === false) {
throw new Error ("Ya need to include a name")
}
this.name = name;
}
}
let user;
try {
user = new User();
} catch (err) {
}
console.log(user); // undefinedBut honestly, who wants to live in a world where we can't trust that a simple new statement won't throw errors. I don't feel like adopting trust issues with my codebase.
We've talked about why throwing errors is not great in the "Functional Error Handling" article, so let's think of a better way to prevent bad objects from being created.
Static factory methods
The best way to enforce validation logic against a domain object is to keep the constructor private and use a static factory method to enforce the constraints.
Using the Result<T> class from "Flexible Error Handling w/ the Result Class | Enterprise Node.js + TypeScript", we can statically represent a success or a failure.
interface UserProps {
name: string;
}
class User {
private props: UserProps;
get name (): string {
return this.props.name;
}
private constructor (props: UserProps) {
this.props = props;
}
public static create (props: UserProps): Result<User> {
const guardResult = Guard.againstNullOrUndefined(props.name, 'name');
const isAppropriateLength = TextUtils.isAtLeast(2, props.name)
&& TextUtils.isAtMost(31, props.name);
// Fail with reason
if (!guardResult.success) {
return Result.fail<User>(guardResult.message)
}
// Fail with reason
if (!isAppropriateLength) {
return Result.fail<User>("Must be between 2 and 31 characters")
}
// Static method can access the constructor
return Result.ok<User>(new User(props));
}
}Now, object creation looks like this:
let user: User;
let userOrError: Result<User> = User.create({ name: 'Khalil Stemmler' });
if (userOrError.isSuccess) {
user = userOrError.getValue();
} else {
console.log(userOrError.error)
}Feel free to get even more functional with these different types of errors as well. We can statically type a NullValue error and an InvalidLength error.
Using the Either<T, U> monad from "Functional Error Handling with Express.js and DDD | Enterprise Node.js + TypeScript", we can build return types like:
type UserResult = Either<
// Failure types
UserErrors.NullValuesError |
UserErrors.InvalidFieldLengthError,
// Success type
User
>Discussion
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2 Comments
How will you solve inheritance?
Why are you Guarding for againstNullOrUndefined on the props, when you have an interface that states that a string value is required for the name property? Is that Guard necessary since Typescript will not let you call the User.create() method without passing in an object with a name property that is set as a string?
Seems like that Guard is unnecessary since the compiler won't let you do that anyway.