Advanced TypeScript argument constraints

Question

I'm wondering if somethings as follows is possible in TypeScript, What I'm trying to achieve:

If the type is inbox then obj should be of type interface IInbox.
If the type is sent then obj should be of type interface ISent.

interface IInbox {

}

interface ISent {

}

class MailClient {
    delete(type: "inbox" | "sent", obj: IInbox | ISent) {

    }
}

let client = new MailClient();
client.delete('inbox', <ISent>{}); // should give compile error

Nitzan Tomer · Accepted Answer · 2016-10-17 18:22:04Z

3

You can define multiple signatures:

class MailClient {
    delete(type: "inbox", obj: IInbox);
    delete(type: "sent", obj: ISent)
    delete(type: "inbox" | "sent", obj: IInbox | ISent) {}
}

But that still won't have a compilation error because your interfaces are identical.
Because typescript is using duck typing then the empty object ({}) satisfies the type requirements.
If you differentiate between the two:

interface IInbox {
    a: string;
}

interface ISent {
    b: string;
}

Then you get your error:

client.delete('inbox', {} as ISent); // Argument of type '"inbox"' is not assignable to parameter of type '"sent"'

(code in playground)

answered Oct 17, 2016 at 18:22

Nitzan Tomer

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Fylax · Accepted Answer · 2016-10-17 18:21:16Z

0

As far as I know this constraint is not possible, anyway you can easily write a very simple workaround as following:

class MailClient {
    delete(type: "inbox" | "sent", objI?: IInbox, objS?: ISent) {
        if ((type === "inbox" && typeof objS !== "undefined") ||
                 type === "send" && typeof objI !== "undefined") {
            throw new "Invalid parameter";    
        }
        let obj = (type === "inbox") ? objI : objS;
    }
}

This would do your job in a quite readable way.

answered Oct 17, 2016 at 18:21

Fylax

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