如何在TypeScript中使用对象解构为构造函数中的类级属性赋值？

class Person {
    name: string;
    age: number;

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

    greet() {
        console.log(`Hello, my name is ${this.name} and I am ${this.age} years old.`);
    }
}

// 使用对象解构创建 Person 实例
const person = new Person({ name: "Alice", age: 30 });
person.greet(); // 输出: Hello, my name is Alice and I am 30 years old.

class Person {
    name: string;
    age: number;

    constructor({ name = "Unknown", age = 0 }: { name?: string; age?: number }) {
        this.name = name;
        this.age = age;
    }

    greet() {
        console.log(`Hello, my name is ${this.name} and I am ${this.age} years old.`);
    }
}

// 使用默认值创建 Person 实例
const personWithDefaults = new Person({});
personWithDefaults.greet(); // 输出: Hello, my name is Unknown and I am 0 years old.