我正在尝试(对我来说)一种新的继承类型。我想使用一个参数来调用继承的函数,该参数确定返回哪个继承者对象。
下面是一个例子来说明我的意思:
function Localizor(type) {
this.language = "English"
this.endonym = "English"
if (window[type]) {
return new window[type]()
}
}
Localizor.prototype.native = function native() {
return "I speak " + this.endonym
}
Localizor.prototype.english = function () {
return "I speak " + this.language
}
function French () {
this.language = "French";
this.endonym = "français"
}
French.prototype = new Localizor()
French.prototype.native = function french() {
return "Je parle " + this.endonym
}
function Thai () {
this.language = "Thai";
this.endonym = "ไทย"
}
Thai.prototype = new Localizor()
Thai.prototype.native = function thai() {
return "พูดภาษา" + this.endonym
}
如果我在不带参数(或使用无效参数)的情况下调用 new Localizor()
,我会得到一个简单的英语对象。如果我用“法语”或“泰语”参数调用它,我会得到一个对象,其中继承者会覆盖一些继承的方法,以便它说法语或泰语。例如:
var thai = new Localizor("Thai")
var feedback = thai.language + " | " + thai.endonym + " | " + thai.english() + " | " + thai.native()
console.log(feedback)
这给了我输出Thai | ไทย |我说泰语 | พูดภาษาไทย
。
我有三个问题:
1. 这种类型的继承是否已在某处记录(是否有名称)?
2、这样工作有危险吗?
3. 此示例检查 window[type]
是否存在,这在浏览器中工作时很有用。如果这是在 Node.js 的模块中,是否有等效的方法来确定模块中是否存在函数?
编辑响应 Zero21xxx
这是我发现的一种检测模块中构造函数是否存在的方法,但对我来说它看起来是危险的乱伦。它有什么风险?有哪些更好的方法可用?
function extend(Child, Parent) {
function F() {}
F.prototype = Parent.prototype
Child.prototype = new F()
//Child.prototype.constructor = Child
Child.parent = Parent.prototype
}
function Localizor(type) {
this.language = "English"
this.endonym = "English"
this.French = function français () {
this.language = "French";
this.endonym = "français"
}
extend(this.French, this)
this.French.prototype.native = function french() {
return "Je parle " + this.endonym
}
this.Thai = function ไทย () {
this.language = "Thai";
this.endonym = "ไทย"
}
extend(this.Thai, this)
this.Thai.prototype.native = function thai() {
return "พูดภาษา" + this.endonym
}
if (typeof this[type] === "function") {
return new this[type]()
}
}
Localizor.prototype.native = function native() {
return "I speak " + this.endonym
}
Localizor.prototype.english = function () {
return "I speak " + this.language
}
module.exports = Localizor
最佳答案
以我的拙见,您应该为 English
、French
和 Thai
提供三个独立的构造函数,它们继承自一个公共(public)构造函数(我们称之为区域设置)。它看起来如下:
function Locale(constructor, language, endonym, native) {
this.constructor = constructor;
this.language = language;
this.endonym = endonym;
this.native = function () {
return native + this.endonym;
};
}
Locale.prototype.english = function () {
return "I speak " + this.language;
};
function English() {}
function French() {}
function Thai() {}
English.prototype = new Locale(English, "English", "English", "I speak ");
French.prototype = new Locale(French, "French", "français", "Je parle ");
Thai.prototype = new Locale(Thai, "Thai", "ไทย", "พูดภาษา");
这导致 separation of concerns :每个构造函数仅精确执行其预期目的。不多不少。现在您可以创建一个 localizer
函数,如下所示:
function localizer(language) {
switch (language) {
case "French": return new French;
case "Thai": return new Thai;
default: return new English;
}
}
因此,您需要做的就是调用localizer
来获取所需的Locale
。
关于JavaScript 通过调用祖先函数进行继承,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/18558324/