calling generic function declared at base class with union type cause unexpected compile error

[wenxiao]

Bug Report

Calling generic function declared at base class with union type cause unexpected compile error.

🔎 Search Terms

Union generic, keyof this

🕗 Version & Regression Information

TypeScript version: 4.7.4

⏯ Playground Link

Playground link with relevant code

💻 Code

class BaseClass {
    func<K extends keyof this>(k: K) {}
}

class A extends BaseClass {
    a = 1;
    b = 2;
}

class B extends BaseClass {
    b = 3;
    c = 4;
}

function func<T, K extends keyof T>(v: T, k: K) {}

let a = new A();
let ab: A | B = new A() as any;

func(a, 'a');
func(a, 'b');
// func(a, 'c'); // compile error

// func(ab, 'a'); // compile error
func(ab, 'b');

a.func('a');
a.func('b');
// a.func('c'); // compile error

// ab.func('a'); // compile error
ab.func('b'); // compile error NOT EXPECTED

🙁 Actual behavior

Compiler Errors

This expression is not callable.
Each member of the union type '((k: K) => void) | ((k: K) => void)' has signatures, but none of those signatures are compatible with each other.

🙂 Expected behavior

last line should not raise Compiler Errors, (<K extends keyof A>(k: K) => void) | (<K extends keyof B>(k: K) => void) should be union to (<K extends keyof A|B>(k: K) => void) ?

Also notice that global function "func" can work as expected.

[RyanCavanaugh]

Duplicate #33591, others.

last line should not raise Compiler Errors, ((k: K) => void) | ((k: K) => void) should be union to (<K extends keyof A|B>(k: K) => void) ?

In general this is not a safe reduction step: contravariant uses of K would become unsound. It's very rarely possible to safely merge two arbitrary signatures.

[wenxiao]

@RyanCavanaugh I found a workaround here:

class BaseClass {
    func<K extends keyof this>(k: K) {}
}

class A extends BaseClass {
    a = 1;
    b = 2;
}

class B extends BaseClass {
    b = 3;
    c = 4;
}

function func<T, K extends keyof T>(v: T, k: K) {}

function workaroundFunc<T extends BaseClass, K extends keyof T>(v:T, k:K){
    v.func(k)
}

let a = new A();
let ab: A | B = new A() as any;

func(a, 'a');
func(a, 'b');
// func(a, 'c'); // compile error

// func(ab, 'a'); // compile error
func(ab, 'b');

a.func('a');
a.func('b');
// a.func('c'); // compile error

// ab.func('a'); // compile error
// ab.func('b'); // compile error NOT EXPECTED
// workaroundFunc(ab, 'a'); // compile error
workaroundFunc(ab, 'b');

Playground link with relevant code

by using workaroundFunc, I can do what I expected to do. I'm not sure what's the different between workaroundFunc and direct call.

[typescript-bot]

This issue has been marked as a 'Duplicate' and has seen no recent activity. It has been automatically closed for house-keeping purposes.