apply higher order type to each type within an union type

[unional]

Search Terms

pick, union, transitive, type alias

Suggestion

Provide a way to define higher order type alias (such as Pick<>) which can apply the type transformation on each type with an union type, instead of combining them together into one type.

Use Cases

Using Pick<> or Omit<> on Discriminated Union types cause the result type no longer a discriminated union.

Examples

type Action = InvokeAction | ReturnAction

type InvokeAction = {
  type: 'invoke',
  id: string,
  payload: string[]
}

type ReturnAction = {
  type: 'return',
  id: string,
  payload: string
}

type Omit<T, K extends keyof T> = Pick<T, Exclude<keyof T, K>>

let actions: Action[] =[]
function push(action: Omit<Action, 'id'>) {
  actions.push(action) // <-- error 'invoke' cannot assign to 'return'
}

The problem is Omit<Action, 'id'> creates this type:

{
  type: 'invoke' | 'return',
  payload: string | string[]
}

instead of

{  type: 'invoke', payload: string[] } | { type: 'return', payload: string }

Using Pick<> has the same problem, but use Omit<> in the example for a cleaner use-case.

Checklist

My suggestion meets these guidelines:

• This wouldn't be a breaking change in existing TypeScript/JavaScript code
• This wouldn't change the runtime behavior of existing JavaScript code
• This could be implemented without emitting different JS based on the types of the expressions
• This isn't a runtime feature (e.g. library functionality, non-ECMAScript syntax with JavaScript output, etc.)
• This feature would agree with the rest of TypeScript's Design Goals.

[jack-williams]

Duplicate of #28339

[unional]

Thanks, it is.

[unional]

However, this issue demonstrate that it could break discriminated unions.

@RyanCavanaugh can you comment on this before I close it again? thx 🌷

[unional]

dup of #31501