Create array with an additional property

Question

I'm using an existing 3rd party API over which I have no control and which requires me to pass an array with an additional property. Something like the following:

type SomeArgument = string[] & { foo: string };
doSomething (argument: SomeArgument);

Now I find it quite clumsy and verbose, to create SomeArgument while keeping the compiler satisfied.

This one works, but no type safety at all:

const customArray: any = ['baz'];
customArray.foo = 'bar';
doSomething(customArray);

Another option which feels cleaner, but is quite verbose (I will need different subclasses with different properties) is to subclass Array:

class SomeArgumentImpl extends Array<string> {
  constructor (public foo: string, content?: Array<string>) {
    super(...content);
  }
}
doSomething(new SomeArgumentImpl('bar', ['baz']));

Is there any better, one-liner-style way? I was hoping for something along doSomething({ ...['baz'], foo: 'bar' }); (this one does not work, obviously).

Dániel Kis-Nagy · Accepted Answer · 2018-09-13 17:15:07Z

1

Suppose this is the function you want to call:

function doSomething (argument: string[] & { foo: string }) {
   argument.push("a");
   argument.foo = "4";
   console.log(argument);
}

Then you can call it like this:

// Works
doSomething(Object.assign(["a", "b"], { foo: "c" }));
// Error
doSomething(Object.assign(["a", 2], { foo: "c" }));
// Error
doSomething(Object.assign(["a", 2], { foo: 4 }));

edited Sep 13, 2018 at 17:15

answered Sep 13, 2018 at 17:03

Dániel Kis-Nagy

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3 Comments

qqilihq Over a year ago

Not quite what I'm looking for: doSomething is actually a 3rd party library function over which I have no control. I just need to provide an argument of type string[] & { foo: string }, which I want to create as simply as possible. Thanks anyway!

Dániel Kis-Nagy Over a year ago

Oh, I thought you're wrapping the library function with a function of your own. I see the problem now :) -- updated my answer.

qqilihq Over a year ago

Ahhh, the idea with Object.assign didn't come to my mind. Perfect!

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