Function overload breaks generics

[OliverJAsh]

TypeScript Version: 3.1.6

Search Terms: overload function generic

Code

declare function identity <T extends unknown>(t: T): T
declare function identity (): undefined

declare class Option<A> {
 chain<B>(f: (a: A) => Option<B>): Option<B>;
}

// Actual type: <T>(option: Option<Option<T>>) => Option<{}>
// Expected type: <T>(option: Option<Option<T>>) => Option<T>
// Disabling second `identity` overload fixes it for some reason.
const optionFlatten = <T>(option: Option<Option<T>>) => option.chain(identity)

[wKich]

I have same issue in another example (TypeScript version: 3.2.2)

function foo(a: number, b: string): void;
function foo(a: string): void;

function foo(a: string | number, b?: string) {}

foo.call(null, 1, "2"); // Expected 2 arguments, but got 3.
foo.bind(null)(1, "2"); // Expected 1 arguments, but got 2.
foo.apply(null, [1, "2"]); // Type 'number' is not assignable to type 'string'.

foo.call(null, "2"); // no errors
foo(1, "2"); // no errors
foo("2"); // no errors

So I can't use call/bind/apply with overloaded functions.

Update: Sorry, I didn't see #27028 (comment)

[Bludator]

I have yet another example:

function Wrap<O extends (...any) => any>(a: O, params: Parameters<O>): void { }

let Bar: {
 Foo<K extends "FooBar">(a: K, b: number): void;
 //comment out this ↓ line:
 Foo(a: never): never;
};

//This should work:
Wrap(Bar.Foo, ["FooBar", 78])

link to playground.

[jack-williams]

Bludator That is a separate issue linked to the design of conditional type inference. When inferring from an overloaded function the last overload is always picked, so Parameters just picks up the last case which happens to be never.

See #21496

[RyanCavanaugh]

Inference only works with the last overload