declare typescript type on spread results

Question

i have code like so:

type myType = {
  key1: string,
  optKey?: boolean,
}

const obj = {
  key1: 'hi',
  key2: 'yes',
}

const { key2, ...typedObj } = obj
typedObj.optKey = true

now typescript complains: Property 'optKey' does not exist on type '{ key1: string; }' Property 'optKey' does not exist on type '{ key1: string; }'

i'd like to inform typescript that typedObj is in fact an instance of myType . is there a way to do this, either during the const spread declaration, or afterwards?

this works:

typedObj2 = typedObj as myType

but is there another way without a temporary untyped variable?

Answer 1

You need to define your obj at first, something that has to do with myType

Here is an example that fixes the problem.

type myType = {
  key1: string,
  optKey?: boolean,
}

const obj: myType & { key2: string } = {
  key1: 'hi',
  key2: 'yes',
}

const { key2, ...typedObj } = obj
typedObj.optKey = true

playground