[英]Why do these all these Javascript prototype inheritance methods seem to work the same?
我正在學習Javascript中的繼承,特別是:Web開發人員專業JS的寄生組合繼承 。 我有3種將SuperType繼承為子類型的方法, 它們的行為都完全相同。 為什么? 應該嗎 我的腸子告訴我我缺少了一些東西
function inheritPrototype(subType, superType) {
var prototype = object(superType.prototype);
prototype.constructor = subType;
subType.prototype = prototype;
}
function myInheritPrototype(subType, superType) {
subType.prototype = Object.create(superType.prototype); // inherit methods
subType.prototype.constructor = subType; // assign constructor
}
function myInheritPrototype2(subType, superType) {
subType.prototype = superType.prototype; // inherit methods
subType.prototype.constructor = subType; // assign constructor
}
這是一個輔助函數:
function object(o) {
function F() {};
F.prototype = o;
return new F();
}
這是使用上面3個InheritPrototype()函數的代碼:
function SuperType1(name) {
this.name = name;
this.colors = ["r", "g", "b"];
}
SuperType.prototype.sayName = function() {
console.log(this.name);
}
function SubType(name, age) {
SuperType.call(this, name); // inherit properties
this.age = age;
}
// method inheritance happens only once
inheritPrototype(SubType, SuperType); // works
//myInheritPrototype(SubType, SuperType); // works
//myInheritPrototype2(SubType, SuperType); // also works, but should it?
SubType.prototype.sayAge = function() {
console.log(this.age);
}
乍一看,第一個和第二個基本相同( object
== Object.create
)。 在第三個函數中, subType.prototype
和superType.prototype
是同一對象,因此SuperType的實例也將是SubType的實例:
function SuperType() {}
function SubType() {}
myInheritPrototype2(SubType, SuperType);
console.log(new SuperType() instanceof SubType) // true
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