Description
TypeScript Version: 3.9.2
Search Terms:
typescript mixin method is any
Code
type AnyCtor = new (...a: any[]) => any
function Foo<T extends AnyCtor>(Base: T) {
return class Foo extends Base {
foo() {}
}
}
function Bar<T extends AnyCtor>(Base: T) {
return class Bar extends Base {
bar() {}
}
}
function One<T extends AnyCtor>(Base: T) {
return class One extends Base {
one() {}
}
}
function Two<T extends AnyCtor>(Base: T) {
return class Two extends Base {
two() {}
}
}
function Three<T extends AnyCtor>(Base: T) {
return class Three extends One(Two(Base)) {
three() {}
}
}
class MyClass extends Three(Foo(Bar(Object))) {
test() {
// @ts-expect-error
this.foo(123)
// @ts-expect-error
this.bar(123)
// @ts-expect-error // ERROR
this.one(123) // this.one is type `any`!
// @ts-expect-error // ERROR
this.two(123) // this.two is type `any`!
// @ts-expect-error
this.three(123)
console.log('no runtime errors')
}
}
const m = new MyClass()
m.test()
Expected behavior:
There should be an error on all the lines marked with // @ts-expect-error
Actual behavior:
There is no type error on the lines with this.one
and this.two
because this.one
and this.two
are seen as type any
.
The expectation is that the one
and two
methods have the proper type (with zero parameters) which would therefore cause a type error from passing in arguments.
Related Issues:
Someone from the Discord chat thought perhaps #29571 might be related, but not with 100% certainty. They do however think this is a bug.
Ultimately, making mixins in TypeScript is too hard. It's too easy to get them wrong and they become a very inconvenient in TypeScript (whereas they are very convenient in plain JavaScript).
Also related: #32080