Why doesn't this code cause a TypeScript type error?

Question

With interfaces defined like so:

interface IRemoteService {
  createRecord(record: RecordType): ng.IPromise<ICreateResponse<string>>;
}

interface ICreateResponse<T> {
  createdId: T;
}

Why doesn't the following code cause a Typescript compilation error?

class RemoteServiceMock implements IRemoteService {
  public static $inject = ["$q"];

  constructor(private $q: ng.IQService){
  }

  createRecord(record: RecordType): ng.IPromise<ICreateResponse<string>> {
    return this.$q.when({});
  }
}

The type of $q.when is when<T>(value: T): IPromise<T>.

Yes... "{}" is of type "Any" because "Any" is not a type... in that it is "any type". — BenjaminPaul
– BenjaminPaul, Commented Jun 3, 2015 at 15:21
Yes... but any type can be of type "any". Its the same as saying... "anything".... which "object" also satisfies... obviously. — BenjaminPaul
– BenjaminPaul, Commented Jun 3, 2015 at 15:27
What I meant by that was "Is {} treated like <any>{}", but you're right, that's probably not clear. Edited. — David Gregory
– David Gregory, Commented Jun 3, 2015 at 15:34

basarat · Accepted Answer · 2015-06-05 02:45:04Z

3

This is according to spec. Here is your example simplified:

interface A{
}
interface B {
  createdId: string;
}

var foo:ng.IPromise<A>;
var bar:ng.IPromise<B>;
bar = foo; // No error

This assignment is allowed if A is a subtype of B or B is a subtype of A. If this is not the case you will get an error as shown below:

interface A {
  breakTypeCompat: number;
}
interface B {
  createdId: string;
}

var foo:ng.IPromise<A>;
var bar:ng.IPromise<B>;
bar = foo; // Error

The reason is the bivariance compatibility of function arguments. See this link for docs + reason why this is the way it is: https://github.com/Microsoft/TypeScript/wiki/Type-Compatibility#function-argument-bivariance

Details

Background

Type compatibility of the interfaces depends upon how you use them. E.g. the following is not an error :

interface IPromise<T>{  
}

interface A{
}
interface B {
  createdId: string;
}

var foo:IPromise<A>;
var bar:IPromise<B>;
bar = foo; // No error

However if the IPromise where to use the type parameter as a member it would error:

interface IPromise<T>{
    member:T    
}

interface A{    
}
interface B {
  createdId: string;
}

var foo:IPromise<A>;
var bar:IPromise<B>;
bar = foo; // Error

Therefore

In the actual promise definition we have something like:

interface IPromise<T> {
    then(successCallback: (promiseValue: T) => any): any;
}

interface A {
}
interface B {
    createdId: string;
}

var foo: IPromise<A>;
var bar: IPromise<B>;
bar = foo; // No Error

Since we are using T as an argument to the a function A and B will be type checked by bivariance. So if A is a subset of B or B is a subset of A they are compatible.

edited Jun 5, 2015 at 2:45

answered Jun 4, 2015 at 1:17

basarat

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2 Comments

David Gregory Over a year ago

The example above certainly works. Sorry to ask a question on top of a question, but why is that allowed? A does not implement B, so that would seem to damage the usefulness of interface membership quite a lot.

David Gregory Over a year ago

Thanks! Makes much more sense now. :)

Brocco · Accepted Answer · 2015-06-03 16:44:32Z

1

I'm not certain why you're not getting an error, but I do have a suggestion on how to get a warning. According to angular.d.ts when is defined like this:

when<T>(value: IPromise<T>): IPromise<T>;
when<T>(value: T): IPromise<T>;
when(): IPromise<void>;

So if you'd like to use when with more typing then use:

return this.$q.when<ICreateResponse<string>>({});

answered Jun 3, 2015 at 16:44

Brocco

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1 Comment

David Gregory Over a year ago

Thanks, this did actually give me an error which is great.

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