Specify arguments of Typescript function callback

Question

I have a callback currently defined like so:

onShow: Function;

And is called like so:

if (this.onShow) {
    this.onShow({ $currentPosition: this.currentLatLngPosition });
}

The main issue with this is that its not typesafe. How can I change the signature of onShow to have better type safety?

I found out I could define it a little better like so

onShow: () => void;

But I could not figure out how to get the correct function arguments in there. This doesnt work:

onShow: ({ $currentPosition: google.maps.LatLng }) => void;

Note the weird json parameter syntax is an Angular component callback requirement.

You can find more information about defining function signatures here: typescriptlang.org/docs/handbook/functions.html — Sebastian Sebald
– Sebastian Sebald, Commented May 10, 2017 at 9:34

Sebastian Sebald · Accepted Answer · 2017-05-10 09:38:53Z

3

interface Config {
    $currentPosition: string; // Or whatever the type of the property is.
}

type myFunc = (config: Config) => void;

You do not have to create an interface:

type myFunc = (config: { $currentPosition:string }) => void;

answered May 10, 2017 at 9:38

Sebastian Sebald

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