Is there a way to extract keys of type A and transfer them to type B?
Even though I anticipate type B to be "x", it seems to also include "undefined".
Why does the keyof operator incorporate undefined in the resulting type? It's perplexing.
I know I can use the Required utility as a workaround, but I'm curious about the inner workings of the keyof operator because the documentation doesn't offer much insight.
playground
type A = {x?: string | null};
type B = {[Key in keyof A]: Key}[keyof A];A type that is homomorphic mapped (explained in this What does "homomorphic mapped type" mean?) takes the shape of {[K in keyof T]: ⋯T[K]⋯}, keeping intact the optional and/or readonly characteristics of the mapped properties. Optional property goes in, optional property comes out:
type A = { x?: string | null };
// ^? { x?: string | null | undefined }
type B0 = { [Key in keyof A]: Key };
// ^? { x?: "x" | undefined }
Furthermore, optional properties inherently encompass undefined within their scope. When you inspect the type using IntelliSense, you will typically see | undefined included, as visible in the example above. (Assuming --strictNullChecks is enabled but not the --exactOptionalPropertyTypes compiler option.)
Consequently, when you access it to extract values as a union, any optional property results in adding an undefined to the union:
type B = B0[keyof A];
// ^? "x" | undefined
Hence, the presence of | undefined isn't due to keyof specifically, but rather from the | undefined linked to the type being mapped.
As indicated, you have the option to utilize Required or the -? mapping modifier to resolve this issue:
type B = { [K in keyof A]-?: K }[keyof A];
// type B = "x"
Check out the code on the TypeScript Playground link
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