Optional parameters based on conditional types

Issue

Is it possible to make a function have either mandatory or optional parameters based on conditional types in TypeScript?

This is what I’ve got so far:

const foo = <T extends string | number>(
    first: T,
    second: T extends string ? boolean : undefined
) => undefined;

foo('foo', true); // ok, as intended
foo(2, true); // not ok, as intended
foo(2, undefined); // ok, as intended
foo(2); // compiler error! I want this to be ok

Solution

You can do this in 3.1 using Tuples in rest parameters and spread expressions

const foo = <T extends string | number>(
  first: T, 
  ...a: (T extends string ? [boolean] : [undefined?])
) => undefined;

foo('foo', true); // ok, as intended
foo(2, true); // not ok, as intended
foo(2, undefined); // ok, as intended
foo(2); // ok

But the better way is to use overloads.

function foo2(first: string, second: boolean) : undefined
function foo2(first: number, second?: undefined): undefined
function foo2<T>(first: T, second?: boolean): undefined{
  return undefined
}

foo2('foo', true); // ok, as intended
foo2(2, true); // not ok, as intended
foo2(2, undefined); // ok, as intended
foo2(2); // ok

Answered By – Titian Cernicova-Dragomir

This Answer collected from stackoverflow, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0

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