TypeScript generics - type inference

Question

I'm trying to recreate something in TypeScript that works in C#. In the code below, I was expecting both response1 and response2 to be a Promise<number>, but response1 doesn't seem to infer the generic type correctly.

Is this possible? Or is it just something that TypeScript cannot do?

interface IRequest<TResponse> {}

interface MyRequest extends IRequest<number> {
    id: string;
}

function execute<TResponse>(request: IRequest<TResponse>): Promise<TResponse>{
    return Promise.reject("not implemented");
}

// const response1: Promise<{}>
const response1 = execute(<MyRequest>{
    id: "123"
});

// const response2: Promise<number>
const response2 = execute(<IRequest<number>>{
    id: "123"
});

What is the type restriction for TResponse

Niladri
– Niladri

2018-06-24 05:41:49 +00:00
Commented Jun 24, 2018 at 5:41 — Niladri
– Niladri, Commented Jun 24, 2018 at 5:41

Titian Cernicova-Dragomir · Accepted Answer · 2018-06-24 06:30:34Z

5

Typescript uses structural compatibility for type compatibility, so things like the inheritance list of an interface are not as important, what matters is the structure of the type. In your case the IRequest interface does not use the TResponse type argument at all, so it's actually structurally equivalent to {}, so it will matter little that MyRequest extends this interface. If IRequest were to use the type parameter, inference will work as expected:

interface IRequest<TResponse> {
  _response?: TResponse
}

interface MyRequest extends IRequest<number> {
    id: string;
}

function execute<TResponse>(request: IRequest<TResponse>): Promise<TResponse>{
    return Promise.reject("not implemented");
}

// const response1: Promise<number>
const response1 = execute(<MyRequest>{
    id: "123"
});

// const response2: Promise<number>
const response2 = execute(<IRequest<number>>{
    id: "123"
});

Playground link

answered Jun 24, 2018 at 6:30

Titian Cernicova-Dragomir

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1 Comment

artem Over a year ago

as it says in the FAQ: "In general, you should never have a type parameter which is unused. The type will have unexpected compatibility (as shown here) and will also fail to have proper generic type inference in function calls."

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