TypeScript-类方法修饰符

• 类的方法修饰符和属性的修饰符一样，只不过方法的修饰符是没有 readonly 的

博主假设有这么一个需求: 有一个基类, 所有的子类都需要继承于这个基类, 但是我们不希望别人能够通过基类来创建对象：

错误示例：

class Person {
    name: string;
    age: number;
    gender: string;

    protected constructor(name: string, age: number, gender: string) {
        this.name = name;
        this.age = age;
        this.gender = gender;
    }

    say(): void {
        console.log(`name=${this.name},age=${this.age},gender=${this.gender}`);
    }
}

class Student extends Person {
    constructor(name: string, age: number, gender: string) {
        super(name, age, gender);
    }
}

let p = new Person('BNTang', 18, 'male');
let stu = new Student('zs', 18, 'female');

正确示例：

class Person {
    name: string;
    age: number;
    gender: string;

    protected constructor(name: string, age: number, gender: string) {
        this.name = name;
        this.age = age;
        this.gender = gender;
    }

    say(): void {
        console.log(`name=${this.name},age=${this.age},gender=${this.gender}`);
    }
}

class Student extends Person {
    constructor(name: string, age: number, gender: string) {
        super(name, age, gender);
    }
}

let stu = new Student('zs', 18, 'female');
console.log(stu);