如何调用带参数的继承JavaScript构造函数？ [英] How Do I Call An Inherited JavaScript Constructor With Parameters?

问题描述

问候，

阅读下面的文章后，我有一个问题：
https://developer.mozilla.org/en/Introduction_to_Object-Oriented_JavaScript

在继承示例中，Person构造函数不接受任何参数。

谢谢！

Person 构造函数的逻辑的方法是使用 https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/Function/callrel =nofollow noreferrer> call 或 apply ，例如：

  function Person（gender）{
 this.gender = gender; 
} 
 
 function Student（gender）{
 Person.apply（this，arguments）; 
} 
 Student.prototype = new Person（）; // make Student继承自Person对象
 Student.prototype.constructor = Student; // fix constructor property 
 
 var foo = new Student（'male'）; 
 foo.gender; //male
 foo instanceof Student; // true 
 foo instanceof Person; // true 
  

如果你想阻止执行 Person 构造函数当没有参数调用时（如下面的行： Student.prototype = new Person（）; ），你可以检测它， p>



  function Person（gender）{
 if（arguments.length == 0）return; //不做任何事
 this.gender = gender; 
} 
  




Greetings,

After reading the following article I have a question: https://developer.mozilla.org/en/Introduction_to_Object-Oriented_JavaScript

In the inheritance example, the Person constructor doesn't take any parameters. How would this same example look if I were to add one and call it from the Student constructor?

Thanks!

解决方案

Well, a way that you can re-use the logic of the Person constructor is invoking it with call or apply, for example:

function Person(gender) {
  this.gender = gender;
}

function Student(gender) {
  Person.apply(this, arguments);
}
Student.prototype = new Person(); // make Student inherit from a Person object
Student.prototype.constructor = Student; // fix constructor property

var foo = new Student('male');
foo.gender;             // "male"
foo instanceof Student; // true
foo instanceof Person;  // true

If you want to prevent the execution of the Person constructor when is called without arguments (like in the line: Student.prototype = new Person();), you can detect it, e.g.:

function Person(gender) {
  if (arguments.length == 0) return; // don't do anything
  this.gender = gender;
}

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