使用属性创建子类，该属性是父类的对应属性的子类

Question

class TableController {
    constructor(public x,public y){}
}
class Table  {
    public controller = TableController;
}

class SquareTableController extends TableController{
  constructor(public x, public y, public z){
      super(x,y);
      /* do other stuff with z */
  }
}
class SquareTable extends Table{
    public controller = SquareTableController;
}

TypeScript为上面的代码提供了以下错误：

TS2415: Class "SquareTable" incorrectly extends base class "Table". 
  Types of property "controller' are incompatible. 
  Type 'typeof SquareTableController' is not assignable to type 'typeof TableController'.

注意SquareTableController在其构造函数中有一个额外的参数。

如何在TypeScript中进行这样的继承设置？我很确定我在C＃和Java中做过非常相似的事情。

可用的游乐场here

Answer 1

用于类型注释的冒号:。

您需要将冒号:用于类型注释，而不是等号=。

class TableController {
    constructor(public x, public y) {}
}

class Table {
    // type annotation
    public controller: TableController;
}

class SquareTableController extends TableController {
  constructor(public x, public y, public z) {
      super(x, y);
  }
}

// type annotation
class SquareTable extends Table {
    public controller: SquareTableController;
}

等于=和new用于对象分配。

如果目标是进行赋值而不是类型注释，那么我们可以使用等号来调用构造函数。

class TableController {
    constructor(public x, public y) {}
}

class Table {
    // object assignment
    public controller = new TableController(1, 2);
}

class SquareTableController extends TableController {
  constructor(public x, public y, public z) {
      super(x, y);
  }
}

class SquareTable extends Table {
    // object assignment
    public controller = new SquareTableController(1,2,3);
}