Is there a way to prevent TypeScript from flagging static members on a function as errors?

In my example, I have outlined the issue at hand.

Essentially, I am seeking to comprehend the proper way to write valid TypeScript code when adding static members to a function in the following code snippet:

export const ZoomPanelControl = (props) => undefined;
ZoomPanelControl.Item = (props) => undefined;

Here is the link to the corresponding code on TypeScript Playground

Answer №1

If you're looking to include a property within a function, one efficient method is to utilize Object.assign() to merge the function and property into a single entity:

const DashboardWidget = Object.assign(
    (props) => undefined,
    { Content: (props) => undefined }
);

By doing this, you create an object that combines the function with the added property, resulting in a new type:

((props: any) => any) & { Content: (props: any) => any; }
, giving you the desired outcome:

DashboardWidget(123); // works fine
DashboardWidget.Content(456); // works fine

Best of luck with your coding endeavors!

Answer №2

If you want to create a separation between defining a function and assigning properties, you can utilize a type assertion:

type Example = {
    (props: any): any;
    [key: string]: (props: any) => any;
}

const ZoomPanelControl = <Example>((props) => undefined);

//...

ZoomPanelControl.Item = (props) => undefined;

While you could embed the type within the type assertion, it can make the code harder to read, which is why I extracted it into the Example type.

The compiler will still validate ((props) => undefined) against the signature of the Example type, ensuring type safety compared to using:

const ZoomPanelControl: Example = <any>((props) => undefined);

Additionally, you can restrict it to specific members if you don't want to allow "any string as a key"...

type Example = {
    (props: any): any;
    Item: (props: any) => any;
}

If you are setting this up all at once, jcalz answer provides type inference benefits (effectively giving you the type in my last code block, but automatically).

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