Typescript 扩展抽象类的类的执行方法

我在Typescript网站上遇到了以下示例：

abstract class Department {

    constructor(public name: string) {
    }

    printName(): void {
        console.log("Department name: " + this.name);
    }

    abstract printMeeting(): void; // must be implemented in derived classes
}

class AccountingDepartment extends Department {

    constructor() {
        super("Accounting and Auditing"); // constructors in derived classes must call super()
    }

    printMeeting(): void {
        console.log("The Accounting Department meets each Monday at 10am.");
    }

    generateReports(): void {
        console.log("Generating accounting reports...");
    }
}

let department: Department; // ok to create a reference to an abstract type
department = new Department(); // error: cannot create an instance of an abstract class
department = new AccountingDepartment(); // ok to create and assign a non-abstract subclass
department.printName();
department.printMeeting();
**department.generateReports();** // error: method doesn't exist on declared abstract type???

关于这一点，我有两个问题：

为什么这一行无效？department可以访问generateReports（），因为它的类型也是会计部门。我已经运行了这段代码，它确实可以毫无问题地执行department.generateReports（）

当我在做

department=newdepartment（）时虽然创建抽象类的实例是一个错误，但它并没有给我一个运行时错误。为什么会这样？不创建抽象类的实例是我的责任吗

部门

被声明为一个

部门

，这意味着即使它可能是一个

会计部门

，也不能保证它是。想想看：

let department: Department;
// ...
department = departmentName === 'accounting'
  ? new AccountingDepartment()
  : new CustomerCareDepartment();
// ...
department.generateReports(); // You really can't know if it's available

如果您只是键入

让department=newaccountingdepartment（）本来可以，因为TS推断类型并将
部门
声明为
会计部门

抽象类是TypeScript特性，而不是JS，因此当TS被传输到JS时，

Abstract

关键字就被删除了。您将在编译时得到错误，就像大多数TypeScript特性一样，而不是在编译时。请参见此处的透明示例：