Void parameter still required when type extends generic

TypeScript Version: 3.2.2

Search Terms: void parameter type extends generic Code

function works1(n: number, b: void) { }

works1(12);

function works2(n: number, b: 1 extends 1 ? void : number) { }

works2(12);

function fails<T>(n: number, b: T extends 1 ? void : number) { }

fails<2>(12, 2);  // works, requires both params
fails<1>(12);  // fails, requires 2 parameters even though second param is void

Expected behavior: That I can ignore the second parameter since its void

Actual behavior: Ts tells me I am missing a parameter

https://www.typescriptlang.org/play/index.html#src=function works1(n%3A number%2C b%3A void) { } works1(12)%3B function works2(n%3A number%2C b%3A 1 extends 1%3F void %3A number) { } works2(12)%3B function fails<T>(n%3A number%2C b%3A T extends 1%3Fvoid%3Anumber) { } fails<2>(12%2C2)%3B %2F%2Fworks%2C requires both params fails<1>(12)%3B %2F%2F requires 2 parameters even though second param is void

[dragomirtitian]

@parzh You can get the parameter names back (mostly) if you don’t mind writing a bit more:

declare function fn<Value extends 1 | 2>(...args: Value extends 1 ?  Parameters<(x: number) =>void> : Parameters<(y:number, name: string) => void>): void;

fn<1>(42);
fn<1>(42, "hello world"); // Error: expects 1 argument

fn<2>(42); // Error: expects 2 arguments
fn<2>(42, "hello world");

Playground Link

[parzh]

Variable number of function parameters can still be achieved using tuple types (although, parameter identifiers in IntelliSense will be lost unfortunately):

declare function fn<Value extends 1 | 2>(...args: Value extends 1 ? [number] : [number, string]): void;

fn<1>(42);
fn<1>(42, "hello world"); // Error: expects 1 argument

fn<2>(42); // Error: expects 2 arguments
fn<2>(42, "hello world");

Link to playground.