4 replies

Question about TypeScript memoization for repeated type utility usage

Hey, does anyone know if TS would memoize

SuperCostlyTypeUtility<X>

SuperCostlyTypeUtility<X>

SuperCostlyTypeUtility<X>

SuperCostlyTypeUtility<X> in the following type? Or would it be paid for twice?

type Foo = <X>(x:X) =>
  (
    & SuperCostlyTypeUtility<X>
    & DoSomething<Y, AndMore<SuperCostlyTypeUtility<X>>>
  )

type Foo = <X>(x:X) =>
  (
    & SuperCostlyTypeUtility<X>
    & DoSomething<Y, AndMore<SuperCostlyTypeUtility<X>>>
  )

type Foo = <X>(x:X) =>
  (
    & SuperCostlyTypeUtility<X>
    & DoSomething<Y, AndMore<SuperCostlyTypeUtility<X>>>
  )

type Foo = <X>(x:X) =>
  (
    & SuperCostlyTypeUtility<X>
    & DoSomething<Y, AndMore<SuperCostlyTypeUtility<X>>>
  )

I'm trying to get TS to "simplify" (compute) the return type and I am limited in what I can do because

DoSomething

DoSomething

DoSomething

DoSomething could return

null

null

so simple

& {}

& {}

& {}

& {} doesn't work here. I did find a solution, but it currently requires a double use of

SuperCostlyTypeUtility

SuperCostlyTypeUtility

SuperCostlyTypeUtility

SuperCostlyTypeUtility