《JavaScript高级程序设计》阅读笔记：面向对象之继承

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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SuperType

SubType.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SuperType

SubType.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"

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

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();//29

var instance2=new SubType("Jack",28);

alert(instance2.colors);//"red","blue","green"

instance2.sayName();//"Jack"

instance2.sayAge();//28

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();//29

var instance2=new SubType("Jack",28);

alert(instance2.colors);//"red","blue","green"

instance2.sayName();//"Jack"

instance2.sayAge();//28