Subclass method is not allowed (TS2425) if the parent class has a property of type any with the same name

TypeScript appears to treat class methods as something fundamentally different from a class prototype property of type function. This is not in agreement with the standard semantics of ES6.

TypeScript Version: all versions, as far as I have been able to test.

Search Terms: method member property any parent class subclass

Code

class Super {
    prop: any;
}

class Derived extends Super {
    prop() {}
}

Expected behavior: should compile without a problem. Note that by the ES6 standard, the above definition of Derived is equivalent to the following, which TypeScript compiles without a problem:

class Derived extends Super {}

Derived.prototype.prop = function(){};

Actual behavior:

Error TS2425: Class 'Super' defines instance member property 'prop', but extended class 'Derived' defines it as instance member function.

Playground Link: https://www.typescriptlang.org/play/#src=class%20Super%20%7B%0D%0A%20%20%20%20prop%3A%20any%3B%0D%0A%7D%0D%0A%0D%0Aclass%20Derived%20extends%20Super%20%7B%0D%0A%20%20%20%20prop()%20%7B%7D%0D%0A%7D%0D%0A

Related Issues: I didn't find similar issues, although I started by looking for issues requesting for the ability to disable particular errors (TS2425 in this case). There are several (closed) issues in which such an ability is requested or suggested, but I believe the proper solution would be to bring TypeScript in agreement with ES6. After all, it claims to be a superset!

リンク先の TypeScript Playground で例を再現することから始め、型が any の継承されたプロパティとサブクラスのメソッドとの比較に対して TS2425 を報告するチェックを追跡します。これを issue で説明されている同等のプロトタイプ代入と比較します。クラス定義が TS2425 なしでコンパイルされ、関連する継承チェックが引き続き維持されていれば完了です。