TypeScript: multiple destructuring and type assertion

Question

I'm wondering how you can assert the type of multiple variables when using destructuring in TypeScript.

Consider:

const { OneTypeOfObject, AndAnother } = ProtoBuf.loadProtoFile('./ohYesProtobufs.proto').build('iAmANameSpace');

How can I assert that OneTypeOfObject has a type of OneType and AndAnother has a type of OtherType ?

I can make them all any like this:

const { OneTypeOfObject, AndAnother } = <any>ProtoBuf.loadProtoFile('./ohYesProtobufs.proto').build('iAmANameSpace');

What I'd actually like to do is something like this:

const { <OneType>OneTypeOfObject, <OtherType>AndAnother } = ProtoBuf.loadProtoFile('./ohYesProtobufs.proto').build('iAmANameSpace');

There's clearly been thought about this sort of thing:

https://github.com/Microsoft/TypeScript/commit/7a74d9f8d021f50201765d2170af8d249af90320

But I can't find the answer based on what I've seen....

Update

I tried both of David's approaches but neither worked. This is the code used for approach 2:

interface OneType extends ProtoBuf.ProtoBuf {
  new(): any;
}

interface OtherType extends ProtoBuf.ProtoBuf {
  new(): any;
}

const { <OneType>OneTypeOfObject, <OtherType>AndAnother } = ProtoBuf.loadProtoFile('./ohYesProtobufs.proto').build('iAmANameSpace');

It caused this error:

Type 'ProtoBuf' is not assignable to type '{ OneTypeOfObject: OneType; AndAnother: OtherType; }'. Property 'OneTypeOfObject' is missing in type 'ProtoBuf'.

Answer 1

Desturcturing with types was confusing to me at first as well. There are there ways you could accomplish this.

The easiest have ProtoBuf.loadProtoFile('./ohYesProtobufs.proto') return the appropriate type (which I realize isn't always possible or easy).

OR

const { OneTypeOfObject, AndAnother } = <{ OneTypeOfObject: TypeA; AndAnother : TypeB }> ProtoBuf.loadProtoFile('./ohYesProtobufs.proto');

OR

const { OneTypeOfObject, AndAnother } : { OneTypeOfObject: TypeA; AndAnother : TypeB } = ProtoBuf.loadProtoFile('./ohYesProtobufs.proto');