Typescript function overload depending on interface generic

Issue

Is there a way to declare a method overload with an optional parameter if a class generic is of a certain type? In the below example the goal is to have a class whose run method requires an argument only if the class's generic T is a number. What I've written doesn't work: I think the T in the run declaration is interpreted as a different generic than Test's generic T. Is there a way to accomplish this?

class Test<T extends string | number> {
  run<T extends number>(a: number): void
  run<T extends string>(a?: number): void
  run(a: number): void {
    console.log(a)
  }
}

const a = new Test<number>();
a.run() // This should be an error, as a's generic is a number and thus its run method should require an argument.

const b = new Test<string>();
b.run() // OK, since b's generic is a string.

Solution

A conditionally-typed rest parameter is a useful choice for your method in this case:

TS Playground

class Test <T extends string | number> {
  run (...args: T extends number ? [a: number] : [a?: number]): void {
    const [a] = args;
    console.log(a); // number | undefined
  }
}

const a = new Test<number>();
a.run(42); // ok
a.run(); /*
  ^^^^^
Expected 1 arguments, but got 0.(2554) */

const b = new Test<string>();
b.run(42); // ok
b.run(); // ok

Answered By - jsejcksn