模拟 extjs 底层继承

 // 混合继承 ???function Person(name,age) { ???????this.name = name; ???????this.age = age; ???} ???Person.prototype = { ???????constructor : Person, ???????sayHello : function () { ???????????alert("hello"); ???????} ???} ???function Boy(name,age,sex) { ???????//Person.call(this,name,age); ???????this.sex = sex; ???} ???Boy.prototype = new Person(); ??????????// ?虽然 在 new Person 的时候，并没有传入参数，但是 还是实例化了 Person 模板 ， 只是参数的值默认为 undefined ???var b = new Boy("z3",25,"男"); ???alert(b.name); ?????????// undefined ???alert(b.sex); ??????????// 男

 // 改进方法 ， 模拟 extjs 底层继承实现方式 ???function Person(name,age) { ???????this.name = name; ???????this.age = age; ???} ???Person.prototype = { ???????constructor : Person, ???????sayHello : function () { ???????????alert("hello"); ???????} ???} ???function Boy(name,age,sex) { ???????Boy.superClass.constructor.call(this,name,age); ???????this.sex = sex; ???} ???function extend(sub,sup) { ???????var F = new Function(); ???????F.prototype = sup.prototype; ???????sub.prototype = new F(); ???????sub.prototype.constructor = sub; ???????sub.superClass = sup.prototype; ???????if(sup.prototype.constructor = Object.prototype.constructor){ ???????????sup.prototype.constructor = sup; ???????} ???} ???extend(Boy,Person); ???var b = new Boy("z3",25,"男"); ???alert(b.name); ?????????????// z3