Typescript Class use Interface as type instead of implement

Question

I'm looking for at way to mimick the C# way of using/implementing Interfaces. In short I'm trying to replicate the following code:

interface EBook {
    function read();
}

class EBookReader {

    private $book;

    function __construct(EBook $book) {
        $this->book = $book;
    }

    function read() {
        return $this->book->read();
    }

}

class PDFBook implements EBook {

    function read() {
        return "reading a pdf book.";
    }
}

class MobiBook implements EBook {

    function read() {
        return "reading a mobi book.";
    }
}

Using implements works fine however I cannot mimick the Class EBookReader way of using Ebook as a type.

codepen with my code mockup: http://codepen.io/Ornhoj/pen/gLMELX?editors=0012

Paleo · Accepted Answer · 2016-11-16 06:46:23Z

2

using Ebook as a type

The case is sensitive.

interface IEBook {
    read();
}

class EBookReader {
    book: IEBook;

    constructor(book: IEBook) {
        this.book = book;
    }

    read() {
        this.book.read();
    }

}

class PDFBook implements IEBook {
    read() {
        console.log("reading a pdf book.");
    }
}

class MobiBook implements IEBook {
    read() {
        console.log("reading a mobi book.");
    }
}
var pdf = new PDFBook();
var reader = new EBookReader(pdf);
reader.read();

Test this code in the playground.

answered Nov 16, 2016 at 6:46

Paleo

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1 Comment

Rudi Ørnhøj Over a year ago

Love it when it's a simple case-sensitive error which has kept me at bay for one-and-a-half day. +1 and correct answer given.

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