Question:

Making an argument required if generic interface as no properties

Problem

How can I make a function argument required based on whether or not P has any properties besides onClose.

So:

If type P has properties besides onClose, props needs to be required and equal to Omit<P, 'onClose'>
If type P has no properties besides onClose, props should not be allowed

export interface Closeable {

onClose: () => void

}

export type WindowOpenFunction = <P extends Closeable>(

window: ComponentType<P>,

props: Omit<P, 'onClose'>

) => void

Solution

There's a way to do it by using spread arguments. First we got to find a way to detect if P has no properties besides onClose.

keyof Omit<P, 'onClose'> extends never

means that there's no extra key inside P besides onClose.

Then we can use a type condition to to do the job:

export type WindowOpenFunction = <P extends Closeable>(

window: P,

...args: keyof Omit<P, 'onClose'> extends never ?

[] | [undefined] | [Omit<P, 'onClose'>]: [Omit<P, 'onClose'>]

) => void

If props have no extra props then the list of extra arguments can either be an empty tuple of one argument or a tuple of one argument, otherwise, it needs to be a tuple of one argument.

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