1.原型链继承

让构造函数的原型对象等于另一个类型的实例，利用原型让一个引用类型继承另一个引用类型的属性和方法

function SuperType(){    this.property=true;}SuperType.prototype.getSuperValue=function(){    return this.property;};function SubType(){    this.subProperty=false;}//继承SuperTypeSubType.prototype=new SuperType();SubType.prototype.getSubValue=function(){    return this.subProperty;}var instance=new SubType();alert(instance.getSuperValue());//true

代码示例中，完整原型链如下

原型链继承的问题：父类型引用类型的属性会被所有子类型实例共享，这是不符合预期的

function SuperType(){    this.colors=["red","blue","green"];}function SubType(){}//继承SuperTypeSubType.prototype=new SuperType();var instance1=new SubType();instance1.colors.push("black");alert(instance1.colors);//"red","blue","green","black"var instance2=new SubType();alert(instance2.colors);//"red","blue","green","black"

2.借用构造函数继承

基本思想是在子类型构造函数内部调用超类型构造函数

function SuperType(){    this.colors=["red","blue","green"];}function SubType(){    //继承SuperType    SuperType.call(this);}var instance1=new SubType();instance1.colors.push("black");alert(instance1.colors);//"red","blue","green","black"var instance2=new SubType();alert(instance2.colors);//"red","blue","green"

借用构造函数可以像超类型构造函数传递参数

function SuperType(name){    this.name=name;}function SubType(){    //继承SuperType    SuperType.call(this,"Jim");    this.age=28;}var instance1=new SubType();alert(instance1.name);//"Jim"alert(instance1.age);//28

借用构造函数的问题：不能复用超类型的方法

3.组合继承

使用原型链实现对原型属性和方法的继承，通过借用构造函数实现对实例属性的继承

function SuperType(name){    this.name=name;    this.colors=["red","blue","green"];}SuperType.prototype.sayName=function(){    alert(this.name);};function SubType(name,age){    //继承SuperType    SuperType.call(this,name);    this.age=age;}SubType.prototype=new SuperType();SubType.prototype.sayAge=function(){    alert(this.age);}var instance1=new SubType("Jim",29);instance1.colors.push("black");alert(instance1.colors);//"red","blue","green","black"instance1.sayName();//"Jim"instance1.sayAge();//29var instance2=new SubType("Jack",28);alert(instance2.colors);//"red","blue","green"instance2.sayName();//"Jack"instance2.sayAge();//28

4.寄生组合式继承

寄生组合式继承，解决了组合继承中，两次调用超类型构造函数的问题

function object(o){    function F(){}    F.prototype=o;    return new F();}function inheritPrototype(subType,superType){    var prototype =object(superType.prototype);    prototype.constructor=superType;//原书是prototype.constructor=subType，看书时认为这里应该是superType    subType.prototype=prototype;}function SuperType(name){    this.name=name;    this.colors=["red","blue","green"];}SuperType.prototype.sayName=function(){    alert(this.name);};function SubType(name,age){    //继承SuperType    SuperType.call(this,name);    this.age=age;}inheritPrototype(SubType,SuperType);SubType.prototype.sayAge=function(){    alert(this.age);}var instance1=new SubType("Jim",29);instance1.colors.push("black");alert(instance1.colors);//"red","blue","green","black"instance1.sayName();//"Jim"instance1.sayAge();//29var instance2=new SubType("Jack",28);alert(instance2.colors);//"red","blue","green"instance2.sayName();//"Jack"instance2.sayAge();//28