Question

请考虑以下代码：

＆＃13;

const defclass = prototype => {
    const constructor = prototype.constructor;
    constructor.prototype = prototype;
    return constructor;
};

const Person = defclass({
    constructor: function Person(firstname, lastname) {
        this.firstname = firstname;
        this.lastname  = lastname;
    },
    get fullname() {
        delete this.fullname; // doesn't delete on instances
        return this.fullname = this.firstname + " " + this.lastname;
    }
});

const john = new Person("John", "Doe");
const jane = new Person("Jane", "Doe");

console.log(john.fullname); // John Doe
console.log(jane.fullname); // Jane Doe

＆＃13;

这是有效的，因为this上的属性赋值会影响不存在的setter。

现在，考虑使用ES6类的相同代码：

＆＃13;

class Person {
    constructor(firstname, lastname) {
        this.firstname = firstname;
        this.lastname  = lastname;
    }

    get fullname() {
        delete this.fullname; // doesn't delete on instances
        return this.fullname = this.firstname + " " + this.lastname;
    }
}

const john = new Person("John", "Doe");
const jane = new Person("Jane", "Doe");

console.log(john.fullname); // throws an error because there is no setter
console.log(jane.fullname);

＆＃13;

this answer解释了它不起作用的原因。这是因为我们在原型链中找到了属性并且它没有设置器。那么，当我们使用常规原型时，为什么不会抛出相同的错误？

注意：您可以使用delete关键字删除该行，而不会影响代码的行为。

Answer 1

我做使用第一个代码获得相同的错误：

const defclass = prototype => {
    const constructor = prototype.constructor;
    constructor.prototype = prototype;
    return constructor;
};

const Person = defclass({
    constructor: function Person(firstname, lastname) {
        this.firstname = firstname;
        this.lastname  = lastname;
    },
    get fullname() {
        "use strict";
//      ^^^^^^^^^^^^
        return this.fullname = this.firstname + " " + this.lastname;
    }
});

const john = new Person("John", "Doe");
const jane = new Person("Jane", "Doe");

console.log(john.fullname); // John Doe
console.log(jane.fullname); // Jane Doe

默认情况下，class代码只在strict mode。

在草率模式下，分配不起作用但被忽略，右侧值从getter开始return。再次访问.fullname会再次运行getter。

为什么懒惰的getter工作原型而不是类？

1 个答案: