为什么有必要设置原型构造函数？ [英] Why is it necessary to set the prototype constructor?

问题描述

在有关MDN文章中继承的部分面向对象的Javascript简介 ，我注意到他们设置了prototype.constructor：

In the section about inheritance in the MDN article Introduction to Object Oriented Javascript, I noticed they set the prototype.constructor:

// correct the constructor pointer because it points to Person
Student.prototype.constructor = Student;  

这是否有任何重要意义？是否可以省略它？

Does this serve any important purpose? Is it okay to omit it?

推荐答案

它并不总是必要的，但确实有它的用途。假设我们想在基础 Person 类上创建一个复制方法。像这样：

It's not always necessary, but it does have its uses. Suppose we wanted to make a copy method on the base Person class. Like this:

// define the Person Class  
function Person(name) {
    this.name = name;
}  

Person.prototype.copy = function() {  
    // return new Person(this.name); // just as bad
    return new this.constructor(this.name);
};  

// define the Student class  
function Student(name) {  
    Person.call(this, name);
}  

// inherit Person  
Student.prototype = Object.create(Person.prototype);

现在当我们创建一个新的学生并复制它？

Now what happens when we create a new Student and copy it?

var student1 = new Student("trinth");  
console.log(student1.copy() instanceof Student); // => false

副本不是学生的实例。这是因为（没有明确的检查），我们无法从base类返回 Student 副本。我们只能返回 Person 。但是，如果我们重置了构造函数：

The copy is not an instance of Student. This is because (without explicit checks), we'd have no way to return a Student copy from the "base" class. We can only return a Person. However, if we had reset the constructor:

// correct the constructor pointer because it points to Person  
Student.prototype.constructor = Student;

...然后一切按预期工作：

...then everything works as expected:

var student1 = new Student("trinth");  
console.log(student1.copy() instanceof Student); // => true

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