In my code, I have the following declaration:
const x : A | (A & B)
I am curious if there exists a way to introduce an "optional" version of the & operator so that I can write more succinctly like this:
const x : A &? B
This would indicate that x definitely meets the criteria for A, but also has the possibility of meeting the criteria for B at times.
Don't be fooled by the nonexistent syntax of A &? B; it doesn't exist. But fear not, for we have the power of utility types at our disposal.
You can craft a utility type for any scenario, although the brevity of using just A | (A & B) might prove more succinct than inventing your own lengthy naming convention.
type WithOptional<Required, Optional> = Required | ( Required & Optional )
const x: WithOptional<A, B> = ...
If dealing with SomeLongName and SomeOtherLongName instead of simpler identifiers like A and B, repeating A in A | (A & B) becomes cumbersome. By extracting out A, we ensure that we always possess A alongside B or another entity.
const x: A & (B | {});
const y: SomeLongName & (SomeOtherLongName | {});
The comments sparked discussions about the practical applications of this type. Personally, I find myself utilizing such constructs for destructuring purposes, where A | (A & B) fails to grant access to B properties. Instead, opting for A & Partial<B> asserts that all A properties are mandatory while allowing B properties to adorn as optional elements. These optional properties may be extracted through destructuring but could potentially hold a value of undefined.
const x: A & Partial<B> = { p1: 0, p2: 0 }
// p1 and p2 are `number`, p3 and p4 are `number | undefined`
const { p1, p2, p3, p4 } = x;
This utility type definition also serves the purpose:
type WithOptional<Required, Optional> = Required & Partial<Optional>
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