Assign Array of types to spread operators

Question

Similar to this example

type Arr = [true, false, "string"]

type t = Arr[0] // true

Id like to assign the types of a sprad operators parameters.

// Not correct syntax; Just thought as an example
function func(...param: [...Arr]) {}

So that the function could be called like this

func(true, false, "string")

Of course for this simple example you could just explicitly tell the function its parameters. Though I need this in order to be able to wrap those types each into an object as generic, without knowing them.

class Data<Type> {

}

class Cls<Types extends any[]> {
  constructor(...data: Data<Types[{{{at index of spread operator}}}]>)
}

If this is not possible, a solution based on infering the Types out of the generics would also be appreciated.

Titian Cernicova-Dragomir · Accepted Answer · 2020-01-18 21:28:17Z

1

You don't need to spread the tuple type, the spread is needed just on the parameter:

type Arr = [true, false, "string"]

function func(...param: Arr) { }
func(true, false, "string")
func(true, false, "stringg") // err

Playground Link

See PR for the whole picture.

To the second part of your question, to warp each tuple member in another generic type, you could use a mapped type:


type Promisify<T> = { 
    [P in keyof T]: Promise<T[P]>
}


function func2(...param: Promisify<Arr>) { }
// Same as 
// func2(param_0: Promise<true>, param_1: Promise<false>, param_2: Promise<"string">): void

Playground Link

edited Jan 18, 2020 at 21:28

answered Jan 18, 2020 at 17:09

Titian Cernicova-Dragomir

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4 Comments

MaximilianMairinger Over a year ago

But how would I wrap each tuple value into a other type as a generic?

Titian Cernicova-Dragomir Over a year ago

@zzrv edited the answer, you just need to use a mapped type

MaximilianMairinger Over a year ago

Thanks alot, ill try this asap

MaximilianMairinger Over a year ago

Your promisify solution workes well for simple cases, though as demonstrated here tinyurl.com/[[[v6429o2]]] it doesnt cope very well with more complex types. I really need to type information in the more first (not working example). Do you have any ideas of how to make this work, or is this an issue with typescript?

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