How to have anonymous object with function property matching another function's signature?

Question

Consider this:

function doIt(thing: SomeThing, iterations: number): void {
  // ...
}

class ThingProcessor{
  private state: {
    // how to avoid specifying the arguments again?
    action: (thing: SomeThing, iterations: number) => void
  };
}

In this example, I have to specify the doIt function parameters twice. I'd rather indicate that state.action is typeof(doIt), e.g. use the same signature as the doIt function.

Possible duplicate of [How to reuse function signature definition in TypeScript](https://stackoverflow.com/questions/47241357/how-to-reuse-function-signature-definition-in-typescript) — Heretic Monkey, Jun 06 '18 at 18:51

score 1 · Accepted Answer · answered Jun 06 '18 at 18:59

1

You can just use the typeof operator on the function to get it's type

function doIt(thing: SomeThing, iterations: number): void {
    // ...
}

class ThingProcessor {
    private state: {
        // how to avoid specifying the arguments again?
        action: typeof doIt;
    };
}

answered Jun 06 '18 at 18:59

Titian Cernicova-Dragomir

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Wow, I was so close! Thanks! – Josh M. Jun 06 '18 at 19:40

How to have anonymous object with function property matching another function's signature?

1 Answers1