TypeScript: How to return conditional types in function with rest parameter?

Question

I want to make a function with rest parameter which will return conditional types like this:

• null when provided undefined value
• string when provided one number
• array of strings when provided multiple numbers

I was trying to write this function in multiple forms using TypeScript extends syntax but unfortunately none of these was working.

I was also trying to write function where first parameter was a number and the second one was rest parameter but it also didn't work.

This is how working code looks right now except of proper conditional return of types:

import { rem } from "polished";

type UnitValue = number | undefined;

const unit = (...values: UnitValue[]): string | string[] | null => {
  const result: string[] = [];

  values.forEach(value => {
    if (value) {
      result.push(rem(value));
    }
  });

  if (result.length === 1) {
    return result[0];
  }

  if (result.length > 1) {
    return result;
  }

  return null;
};

Recreated case in codesansbox.io -> link

I need this function to return exactly and only these types in those three cases:

• unit(undefined) -> null
• unit(16) -> string
• unit(16, 32) -> string[]

Answer 1

You can achieve this behavior by using function overloads . Your code will look like this:

import { rem } from "polished";

function unit(...values: undefined[]): null;
function unit(...values: [number]): string;
function unit(...values: number[]): string[];
function unit(...values): null | string | string[] {
  const result: string[] = [];

  values.forEach(value => {
    if (value) {
      result.push(rem(value));
    }
  });

  if (result.length === 1) {
    return result[0];
  }

  if (result.length > 1) {
    return result;
  }

  return null;
}

You can check out in this playground .