javascript - 检索原型(prototype)属性时出现类型错误

Question

我有以下代码:

function inheritPrototype (sup, sub) {
  var proto = Object.create(sup.prototype);
  Object.defineProperty(proto, "constructor", {value : sub});
  sub.prototype = proto;
}

function Person (name, age) {
  this.name = name;
  this.age = age;

  if (!Person.prototype.getName) {
    Person.prototype.getName = function () { return this.name; }
    Person.prototype.getAge = function () { return this.age; }
  }
}

function Employee (name, age, skills) {
  Person.call(this, name, age);
  this.skills = skills;

  if (!Employee.prototype.getSkills) {
    inheritPrototype(Person, Employee);
    Employee.prototype.getSkills = function () { return this.skills; }
  }
}

var person = new Person ("Dave", 21);
var employee = new Employee ("David", 22, ["C", "C++", "Java", "Python", "PHP"]);

console.log(employee.getSkills());

inheritPrototype 的定义只是为了防止父级 (Person) 构造函数的双重调用。为了保持清晰，我在构造函数内分配原型(prototype)属性，并为了防止每次创建新实例时发生这种情况，我检查原型(prototype)属性是否存在。如果是这样，那么我们不想重新分配属性，否则我们会这样做。问题是，我收到一个 TypeError，提示“employee.getName 不是函数”。仅当我尝试使用 Employee 实例访问 Employee 的原型(prototype)属性时才会发生这种情况。 person 构造函数具有相同的方法来分配 Prototype 属性，但它工作得很好。

console.log(person.getName()); // "Dave"
console.log(employee.getSkills()); // or getName or anything, TypeError

我想我在那里做了一些愚蠢的事情，但无法发现它。那么，出了什么问题？

Answer 1

在外部调用inheritPrototype怎么样？

这对我有用:

function inheritPrototype (sup, sub) {
  var proto = Object.create(sup.prototype);
  Object.defineProperty(proto, "constructor", {value : sub});
  sub.prototype = proto;
}

function Person (name, age) {
  this.name = name;
  this.age = age;

  if (!Person.prototype.getName) {
    Person.prototype.getName = function () { return this.name; }
    Person.prototype.getAge = function () { return this.age; }
  }
}

function Employee (name, age, skills) {
  Person.call(this, name, age);
  this.skills = skills;

  if (!Employee.prototype.getSkills) {
    Employee.prototype.getSkills = function () { return this.skills; }
  }
}

inheritPrototype(Person, Employee); // <= FIX

var person = new Person ("Dave", 21);
var employee = new Employee ("David", 22, ["C", "C++", "Java", "Python", "PHP"]);

console.log(person.getName()); // returns "Dave"
console.log(employee.getSkills()); // returns ["C", "C++", "Java", "Python", "PHP"]
console.log(employee.getName());  // returns David