Compiler should complain about implicit return value of child class constructor, if it does not satisfy child class type

TypeScript Version:  2.1.5

Since TypeScript 2.1 (due to #7574) the return value of a constructor call is the return value of the super constructor. So that

class Child extends Base {}

is compiled to

function Child() {
     return _super !== null && _super.apply(this, arguments) || this;
}

Actual behavior:
The actual problem is, that the implementation of Child(see example below) will be compiled without errors, despite of the fact that the instance (child) does not satisfy its type Child, because someMethod does not exist on child.

class Base {
    hello = 'base';
    constructor() {
        return {
            hello: 'world'
        };
    }
}

class Child extends Base {  // <-- compiler does not complain
    someMethod() {}
}

const child = new Child();
child.someMethod();  // <-- compiler does NOT complain, despite of "someMethod" does not exist on {hello: 'world'}

Expected behavior:
The compiler should complain about the implicit return value of the child class constructor, if this value is not a type of the child class. So that either an explicitly defined return value for the child class constructor should be forced like...

class Child extends Base { 

   constructor() {
     super();
     return { 
       someMethod() {},
       /* ... inherited members */
     }
   }
   someMethod() {}
}

or the return value has to be extended (const base = super()) to fulfill the type Child or all members of the child class have to be optional.

この issue には TypeScript のインライン再現コードがありますが、ソースファイルやテストは指定されていません。まず派生コンストラクターのケースを再現し、コンパイラーによるコンストラクターの戻り値の型の処理を追跡してください。完了条件は、暗黙の子コンストラクターの戻り値が子クラスの型を満たせない場合にコンパイラーが報告することです。