TypeScript Array Union Type in function parameters

Question

I have a union type of an Array of various specific lengths:

[ number ] | [ number, number ] | [ number, number, number, number ]

As you can see, there are requirements for an array with one element, two elements, or four elements.

I am trying to create an object which contains a function with one of these lengths. How do I write the type definition to allow for this?

TS playground

Example:

const people: {
    name: string,
    address: Address,
    work: (numbers: [ number ] | [ number, number ] | [ number, number, number, number ]) => any
}[] = [
    {
        name: "Bob",
        address: new Address(),
        work: function(numbers: [ number ]): number {
            // Implementation returning number
        }
    },
    {
        name: "Ashley",
        address: new Address(),
        work: function(numbers: [ number, number, number, number ]): boolean {
            // Implementation returning boolean
        }
    },
    {
        name: "Michael",
        address: new Address(),
        work: function(numbers: [ number, number ]): number {
            // Implementation returning number
        }
    },
]

Currently, it's giving me the error:

Type '(numbers: [number]) => number' is not assignable to type '(numbers: [number] | [number, number] | [number, number, number, number]) => any'. Types of parameters 'numbers' and 'numbers' are incompatible. Type '[number] | [number, number] | [number, number, number, number]' is not assignable to type '[number]'. Type '[number, number]' is not assignable to type '[number]'. Source has 2 element(s) but target allows only 1.ts(2322)

------- Edit -------

I've applied a suggestion from the comments and made all the possible calls into separate function unions instead of an array union:

const people: {
    name: string,
    address: Address,
    work: ((numbers: [ number ]) => any) | ((numbers: [ number, number ]) => any) | ((numbers: [ number, number, number, number ]) => any)
}[] = [

When trying to now call a function from this array:

 people[1].work([2, 8, 6, 4])

It throws the following error now:

In VSCode I found out this is why:

"The intersection '[number] & [number, number] & [number, number, number, number]' was reduced to 'never' because property 'length' has conflicting types in some constituents."

Answer 1

UPDATED You need to use bivariance here

class Address { }

type Tuple<
  N extends number,
  Item = number,
  Result extends Array<unknown> = [],
  > =
  (Result['length'] extends N
    ? Result
    : Tuple<N, Item, [...Result, number]>
  )


interface WorkFn {
  work(numbers: Tuple<1> | Tuple<2> | Tuple<4>): any
}

interface Person extends WorkFn {
  name: string,
  address: Address,
}

const people: Person[] = [
  {
    name: "Bob",
    address: new Address(),
    work(numbers: Tuple<1>) {
      const [myNumber] = numbers;

      return myNumber * 6
    }
  },
  {
    name: "Ashley",
    address: new Address(),
    work: function (numbers: Tuple<4>): boolean {
      const [myNumber, anotherNumber, someNumber, replaceNumber] = numbers;

      return myNumber === anotherNumber && someNumber === replaceNumber;
    }
  },
  {
    name: "Michael",
    address: new Address(),
    work: function (numbers: Tuple<2>): number {
      const [myNumber, anotherNumber] = numbers;

      return myNumber * anotherNumber;
    }
  },
]

TypeScript playground

Here you can find the difference between method type an arrow function type and about bivariance

Also, please be aware that it is not 100% safe

Answer 2

My proposal would be to use a generic function to create the people array.

function createPeople<
  T extends {
    name: string,
    address: Address,
    work: ((numbers: [ number ]) => any) | ((numbers: [ number, number ]) => any) | ((numbers: [number, number, number, number]) => any)
  }[]
>(p: [...T]){
  return p
}

const people = createPeople([
  {
    name: "Bob",
    address: new Address(),
    work: function(numbers: [ number ]): number {
      const [ myNumber ]: [ number ] = numbers;

      return myNumber * 6
    }
  },
  {
    name: "Ashley",
    address: new Address(),
    work: function(numbers: [ number, number, number, number ]): boolean {
      const [ myNumber, anotherNumber, someNumber, replaceNumber ]: [ number, number, number, number ]= numbers;

      return myNumber === anotherNumber && someNumber === replaceNumber;
    }
  },
  {
    name: "Michael",
    address: new Address(),
    work: function(numbers: [ number, number ]): number {
      const [ myNumber, anotherNumber ]: [ number, number ] = numbers;

      return myNumber * anotherNumber;
    }
  },
])

TypeScript now knows what the callback of each index is. That makes the following calls strictly typed.

people[1].work([2, 8, 6, 4])
people[2].work([1, 2])

---

Playground