CODEWITHSUNDEEP
typescriptencapsulation
lodo
lodo

Reputation: 2383

Typescript: abstract class and module-level visibility

I saw that in Typescript you can emulate module visibility with interfaces, but I don't know if it is possible to achieve it in the following scenario:

abstract class ConnectionTarget
{
    // callback that subclasses must implement
    protected abstract onConnection: (conn: Connection) => void;

    // property that must be available to subclasses
    protected get connections(): Readonly<Iterable<Connection>>
    {
        return this.conns;
    }

    // private field needed for previous property
    private conns: Connection[] = [];

    // method that SHOULD HAVE MODULE VISIBILITY
    // my module should be able to add connections,
    // but my users shouldn't
    private addConnection(conn: Connection)
    {
        this.conns.push(conn);
        this.onConnection(conn);
    }
}

// my function that needs access to the private members
// the parameter is a user-provided subclass of ConnectionTarget
function doMagicThings(target: ConnectionTarget, source: ConnectionSource)
{
    // do magic tricks here ...

    // method that should be module-protected, like addConnection
    let aConnection: source.createConnection();

    target.addConnection(aConnection);
}

I'd like my users to extend ConnectionTarget, having to implement onConnection and being able to only use the property connections, with everything else hidden to them.

EDIT: example usage

// class in user code
class MyConnectionTarget extends ConnectionTarget
{
    // users must implement this abstract method
    onConnection(conn: Connection)
    {
        // user specific code here
        // ...

        // can use property 'connections'
        console.log(this.connections)

        // should error here:
        // should not allow to use the following method
        this.addConnection(new Connection());
    }
}

Upvotes: 1

Views: 1588

Answers (1)

Nitzan Tomer
Nitzan Tomer

Reputation: 164417

You can do that by exporting an interface which declares the public methods without exporting the class itself.
You will then need a factory function which is exported by the module to be able to instantiate the class, something like:

export interface IConnectionTarget {
    // public methods will be declared here, i.e:
    myMethod(): void;
}

abstract class ConnectionTarget implements IConnectionTarget {
    private conns: Connection[] = [];

    protected abstract onConnection: (conn: Connection) => void;

    protected get connections(): Readonly<Iterable<Connection>> {
        return this.conns;
    }

    public addConnection(conn: Connection) {
        this.conns.push(conn);
        this.onConnection(conn);
    }

    public myMethod() {}
}

export function createConnectionTarget(): IConnectionTarget {
    // create an instance here and return it
}

(code in playground)

Edit

Without understanding what you're trying to do better, at seems that you have a few options, but none of them is very pretty:

(1) Keep the method private and when trying to access it cast to any:

let aConnection: source.createConnection();
(target as any).addConnection(aConnection);

(2) Save the setter in the ctor to a module level store:

type Adder = (conn: Connection) => void;
const CONNECTION_ADDERS = new Map<ConnectionTarget, Adder>();

abstract class ConnectionTarget {
    protected constructor() {
        CONNECTION_ADDERS.set(this, this.addConnection.bind(this));
    }

    private addConnection(conn: Connection) { ... }
}

And then to use it:

let aConnection: source.createConnection();
CONNECTION_ADDERS.get(aConnection)(aConnection);

Upvotes: 1

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