Is there a way to modify an interface by making optional properties required?
Consider the following interface:
interface Foo {
test1: string;
test2?: string;
}
I want to create a new interface based on this, with the optional property now being required:
interface FooDefaults {
test2: string;
}
This new interface can be used for setting default values like so:
const defaults: FooDefaults = {
test2: 'bar'
}
This particular issue is quite complex, as there are ongoing debates about the syntax. You can find more details in this thread. The solution provided below is influenced by @jcalz's response on this post.
interface Bar {
example1: string;
example2?: string;
}
type Defaults<T, OptionalKeys extends keyof T = { [K in keyof T]:
({} extends { [P in K]: T[K] } ? K : never)
}[keyof T]> = {
[K in OptionalKeys]-?: T[K]
}
type BarDefaults = Defaults<Bar>
In my opinion, this solution appears to be more of a workaround than a definitive fix. It relies on the behavior where an empty object {} is considered to extend an object with an optional property ({a?: string}), but not one with a required property ({a: string}).
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