Infer interface type argument in TypeScript

Question

Let's have a following code:

interface Action {
  type: string;
}

interface ActionA extends Action {
  type: 'A';
  payload: number;
}

interface ActionB extends Action {
  type: 'B';
  payload: string;
}

interface Processor<T extends Action> {
  action: T;
  process: (action: T) => void;
}

// usage

const actionA: ActionA = { type: 'A', payload: 42 };

const processorA: Processor<ActionA> = {
  action: actionA,
  process(action) {
    // ...
  },
};

Now I think that specifying type argument ActionA in const processorA: Processor<ActionA> = ... is redundant as it could be inferred from action: ActionA. Unfortunately Typescript reports error if I write just const processorA: Processor = ....

Is it possible to improve interfaces so that type argument of Processor would be inferred?

Advanced version:

I would also like action field to be of type T | '*'. In that case action parameter of process(action) should be of type Action (or in worst case just any). Is this possible together with above mentioned type parameter inferring?

Answer 1

There is no way in typescript to infer part of a variable type. For variables inference is all or nothing, you either let the compiler infer it, or you specify it in a type annotation.

You can use a function to infer the type of the action:

function createProcessor<T extends Action>(p: Processor<T>) {
    return p
}
const processorA = createProcessor({
  action: actionA,
  process(action) {
    // ...
  },
});

Or you can use an IIFE as a kind of fancy type assertion:

const processorA = (<T extends Action>(p: Processor<T>) => p)({
  action: actionA,
  process(action) {
    // ...
  },
});