Conditional typing

[burnt flare]

Just curious if its possible
So for example i have a type

type Props = {
  zoom?:boolean,
  options?:Record<string,any>
}

Is it possible if value of zoom is true then options is present in type and if zoom is false then options would be never?

[simple berry]

use a discriminated union

what if zoom is missing though?

(sidenote; never is different to not being present)

[rose flint]

!hb isc

[ripe garden]

Reference / Symbols

Symbol.isConcatSpreadable

[rose flint]

bruh

!hb disc

[ripe garden]

Get Started / TypeScript for Functional Programmers

Discriminated Unions

[burnt flare]

Im just rendering different templates
If zoom is true then renders a wrapper which has options props and if not then im having different thing
The idea was if i pass zoom prop then i get type suggestion for options and if not i would not get them, its totally not necessary but i was just wondering if its possible to make

[simple berry]

lmao

[rose flint]

bruh what even is this

[simple berry]

so if zoom is missing it's also no options?

[burnt flare]

Well technically yes

I just not sure what way to go

[rose flint]

i mean, thats fine too

[burnt flare]

either have the type all the time

or not

[simple berry]

would be ts type Props = { zoom: true; options: object; } | { zoom?: false; }then

[rose flint]

yeah

[burnt flare]

oh dang

[rose flint]

i actually use this pattern for multiple?: false in some react components

[burnt flare]

never thought about this that way