我编写了一个extend方法来实现javascript中的继承:
function Class() {}
Class.prototype.create = function () {
var instance = new this();
instance.init();
return instance;
}
// extend method
Class.extend = Class.prototype.extend = function (props) {
var SubClass = function () {};
SubClass.prototype = Object.create(this.prototype);
for (var name in props) {
SubClass.prototype[name] = props[name];
}
SubClass.prototype.constructor = SubClass;
if (this.prototype.init) {
SubClass.prototype.callSuper = this.prototype.init;
}
SubClass.extend = SubClass.prototype.extend;
SubClass.create = SubClass.prototype.create;
return SubClass;
}
// level 1 inheritance
var Human = Class.extend({
init: function () {
}
});
// level 2 inheritance
var Man = Human.extend({
init: function () {
this.callSuper();
}
})
// level 3 inheritance
var American = Man.extend({
init: function () {
this.callSuper();
}
})
// initilization
American.create();
然后开发工具报告Maximum call stack size exceeded
我认为callSuper
方法导致问题,callSuper
调用init
和init
调用callSuper
,两者都具有相同的上下文。
任何人都可以帮助我吗?如何设置正确的上下文?
答案 0 :(得分:1)
您有范围问题。这是解决方案:
function Class() {}
Class.prototype.create = function () {
var instance = new this();
instance.init();
return instance;
}
// extend method
Class.extend = Class.prototype.extend = function (props) {
var SubClass = function () {},
self = this;
SubClass.prototype = Object.create(this.prototype);
for (var name in props) {
SubClass.prototype[name] = props[name];
}
SubClass.prototype.constructor = SubClass;
if (this.prototype.init) {
SubClass.prototype.callSuper = function() {
self.prototype.init();
}
}
SubClass.extend = SubClass.prototype.extend;
SubClass.create = SubClass.prototype.create;
return SubClass;
}
// level 1 inheritance
var Human = Class.extend({
init: function () {
console.log("Human");
}
});
// level 2 inheritance
var Man = Human.extend({
init: function () {
console.log("Man");
this.callSuper();
}
})
// level 3 inheritance
var American = Man.extend({
init: function () {
console.log("American");
this.callSuper();
}
})
// initilization
American.create();
关键时刻是将 init 方法包装在一个闭包中:
SubClass.prototype.callSuper = function() {
self.prototype.init();
}
这是一个包含解决方案http://jsfiddle.net/krasimir/vGHUg/6/
的jsfiddle