TypeScript 1.5: Export default implementing an interface

Question

In TypeScript 1.5, I have the following interface in IFoo.ts:

// IFoo.ts
interface IFoo<T> {
    bar(): T;
}

And an implementation in FooString.ts:

// FooString.ts
export default {
    bar: () => 'Hello world'
}

How can the module FooString.ts declare the object literal it is exporting as an implementation of IFoo<sring>? Without a declaration, the implementation of the interface is not checked by the compiler, and losing compile-time checking of the FooString module would be problematic.

Deathspike · Accepted Answer · 2015-04-24 20:59:46Z

2

Casting in 1.5 will preserve the compile-time checking, so this will work:

export default <IFoo<string>> {
    bar: () => 'Hello world'
}

answered Apr 24, 2015 at 20:59

Deathspike

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vasa_c · Accepted Answer · 2015-04-24 16:09:35Z

-1

export var defaults:IFoo<string> = {
    bar: () => 'Hello world'
}

answered Apr 24, 2015 at 16:09

vasa_c

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2 Comments

gung - Reinstate Monica Over a year ago

This is rather sparse. Can you add some text to explain this code / how it solves the OP's problem?

Deathspike Over a year ago

Unfortunately not the appropriate answer. Exporting a variable named defaults does nothing.

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