javascript - JavaScript 对象中的私有(private)变量定义

Question

我正在学习使用原型(prototype)属性在 JavaScript 上定义“类似类”的函数。当我像这样定义一个 People 类时，一切都运行良好:

var People = function(firstName, lastName){
    // public
    this.firstName = firstName;
    this.lastName  = lastName;

    this.fullName = function(){
        console.log(full_name);
        return this;
    };

    // private
    var self = this;
    var full_name = function(){
        return self.firstName + ' ' + self.lastName;
    }();
};

var person = new People('John', 'Smith');
person.fullName();  // => "John Smith"

但是，当我将 fullName 方法移动到初始定义之外时，如下所示:

var People = function(firstName, lastName){
    // public
    this.firstName = firstName;
    this.lastName  = lastName;

    // private
    var self = this;
    var full_name = function(){
        return self.firstName + ' ' + self.lastName;
    }();
};

People.prototype.fullName = function(){
    console.log(full_name);
    return this;
};

var person = new People('John', 'Smith');
person.fullName();  // => "ReferenceError: full_name is not defined"

我遇到了一个错误。我不明白为什么...？？

Answer 1

这里:

var full_name = function(){
    return self.firstName + ' ' + self.lastName;
}();

您将变量“full_name”设置为调用该匿名函数的结果。然而，它只是在该构造函数的上下文中声明的一个变量，所以这并不重要；构造函数之外的任何东西都无法访问它。

因此在该原型(prototype)函数中，您会得到一个错误，因为“full_name”实际上未定义。如果您想在原型(prototype)上使用全名函数，只需在此处声明即可:

People.prototype.fullName = function() {
  return this.firstName + " " + this.lastName;
};

如果您想要一个只能由构造函数中的代码访问的“私有(private)”函数，那么您可以这样做:

var People = function(firstName, lastName){
    // public
    this.firstName = firstName;
    this.lastName  = lastName;

    // private
    var self = this;
    function fullName(){
        return self.firstName + ' ' + self.lastName;
    }

    this.addressMe = function(msg) {
        return fullName() + ": " + msg;
    };
};

然后你可以这样做:

var bob = new People("Bob", "Scum");
alert( bob.addressMe("stand and deliver") ); // "Bob Scum: stand and deliver"