CODEWITHSUNDEEP
typescripttypescript-typings
Carven
Carven

Reputation: 15660

How can I copy the signature of a method to another method?

I'm using a library that has this class:

class Dog {
  public run(speed: number, movement: number): void;
  public run(speed: number, type: string): void;
  public run(speed: number, opts: string | number): void {
    // perform some stuff
  }
}

I want to create another interface with a different method name but with the same method signature as Dog.run. I have tried this but it doesn't work because I cannot refer to the run method in Dog by using Dog.run:

interface SpecialDog {
   runWithSpecialPower: (...args: Parameters<Dog.run>) => ReturnType<Dog.run>
}

In this case, I can't extend SpecialDog from Dog too because I want the method name to be runWithSpecialPower instead of just run. Is there a way I can copy the method signature of another method?

TS Playground

Upvotes: 2

Views: 733

Answers (1)

jcalz
jcalz

Reputation: 329868

In this case you're probably just looking for indexed access types. If you have an object foo of type Foo with a key named "bar", then the type of foo.bar is Foo["bar"], using the bracket notation. (You can't look up property types with dot notation, as it conflicts with namespace/module exports, see microsoft/TypeScript#30815.) In your case, you want Dog["run"]:

interface SpecialDog {
  runWithSpecialPower: Dog["run"];
}

And let's verify that it works:

declare const specialDog: SpecialDog;
specialDog.runWithSpecialPower(1, 2);
specialDog.runWithSpecialPower(1, "okay");

Looks good.

Playground link to code

Upvotes: 2

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