microsoft / microsoft/TypeScript

Covariance breaks when checking for `undefinded` in a type

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#55,161 3 comentarios 0 reacciones 1 asignado Ver en GitHub

@weswigham ya está trabajando en esto.

Desde el 27/7/2023.

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Descripción

Bug Report

I'm using the solid-js npm package which defines a type called Setter:

type Setter<T> = (undefined extends T ? () => undefined : {}) & (<U extends T>(value: (prev: T) => U) => U) & (<U extends T>(value: Exclude<U, Function>) => U) & (<U extends T>(value: Exclude<U, Function> | ((prev: T) => U)) => U);

I don't fully understand the reasoning behind it, but it's giving me some problems:

type Setter<T> = (undefined extends T ? () => undefined : {}) & (<U extends T>(value: (prev: T) => U) => U) & (<U extends T>(value: Exclude<U, Function>) => U) & (<U extends T>(value: Exclude<U, Function> | ((prev: T) => U)) => U);

class A { a = 1 }
class B extends A { b = 2 }

type Test<T> = { 0: Setter<T> };

declare const a: Test<A>;
const b: Test<B> = a; // ERROR!

I created an object with the property 0 which contains a Setter and it appears to not be covariant.
Inspecting the error it looks like that the problem is originated from this part undefined extends T ? () => undefined : {}, infact if I remove it this works, moreover the problem would still occur if that was the only part of the type:

type Setter<T> = undefined extends T ? () => undefined : {};

class A { a = 1 }
class B extends A { b = 2 }

type Test<T> = { 0: Setter<T> };

declare const a: Test<A>;
const b: Test<B> = a; // ERROR!

I think it is an error because the types are the same:

type TA = Test<A>;
//   ^? type TA = { 0: {}; }
type TB = Test<B>;
//   ^? type TB = { 0: {}; }

Additionally if Test were to be defined as [ Setter<T> ] the error would not occur:

type Setter<T> = undefined extends T ? () => undefined : {};

class A { a = 1 }
class B extends A { b = 2 }

type Test<T> = [ Setter<T> ];

declare const a: Test<A>;
const b: Test<B> = a; // OK?!?!
🔎 Search Terms
  • undefined extends T
  • covariance
  • covariant
🕗 Version & Regression Information

This changed between versions 3.3.3 and 3.5.1. (The only playground version in which it doesn't occur is 3.3.3)

⏯ Playground Link

Playground Link

💻 Code
type Setter<T> = (undefined extends T ? () => undefined : {}) & (<U extends T>(value: (prev: T) => U) => U) & (<U extends T>(value: Exclude<U, Function>) => U) & (<U extends T>(value: Exclude<U, Function> | ((prev: T) => U)) => U);
// type Setter<T> = (<U extends T>(value: (prev: T) => U) => U) & (<U extends T>(value: Exclude<U, Function>) => U) & (<U extends T>(value: Exclude<U, Function> | ((prev: T) => U)) => U);
// type Setter<T> = undefined extends T ? () => undefined : {};

class A { a = 1 }
class B extends A { b = 2 }

type Test<T> = { 0: Setter<T> };
// type Test<T> = [ Setter<T> ];

declare const a: Test<A>;
const b: Test<B> = a; // ERROR!

type TA = Test<A>;
//   ^?
type TB = Test<B>;
//   ^?
🙁 Actual behavior

Can't assign b with the value of a.
It is wrong at the very least because it is not consistent

🙂 Expected behavior

Can assign b with the value of a

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