javascript - 奇怪的递归无限循环我无法追踪

Question

我有一个原型(prototype)，其中有一个可以添加回调的方法:

/*
 * Add a callback function that is invoked on every element submitted and must return a data object.
 * May be used as well for transmitting static data.
 * 
 * The callback function is supposed to expect a jQuery element as single parameter 
 * and must return a data object (for additional data to be sent along with the one already given upon initialization).
 * Adding multiple callback functions results in those functions being invoked in the same order as they were added.
 * 
 * 1) Therefore subsequent assignments to the same key in the data array override those that were performed by a previous callback function.
 * 2) If data_arg = true is given, the data returned by the callback function that was previously called is given to the new_callback as 2nd argument, so it can be manipulated.
 *    However, even if it isn't, the unchanged data must be returned anyway to have any effect.
 */

this.add_data_callback = function(new_callback, data_arg) {

    if(this.data_callback) {
        old_callback = this.data_callback;

        if(!data_arg) {
            //alert('add it');
            //alert(old_callback);
            //alert(new_callback);

            this.data_callback = function(element) {
                //alert('do it');
                //alert(old_callback);
                //alert(new_callback);
                return jQuery.extend(old_callback(element), new_callback(element));
            };
        }
        else {
            this.data_callback = function(element, data) {
                return new_callback(element, old_callback(element));
            };
        }
    }
    else {
        //alert('add the first');
        //alert(new_callback);
        this.data_callback = new_callback;
    }
};

(请忽略 data_arg = true 的 else 部分，因为它不相关。)

在我的具体案例中，我添加了三个回调函数。然而，如果最终为一个元素调用 this.data_callback() ，则会导致整个事件无限循环。在我尝试追踪错误时，上面的警报(是的，我知道有调试工具，但这样更舒服)提供了对该问题的以下见解:

1. 调用匿名函数/闭包(包含 jQuery.extend() 的函数/闭包)。new_callback 包含第三个回调函数。old_callback 包含另一个...
2. 被调用的匿名函数。new_callback 包含第二个回调函数。old_callback 应该包含第一个回调函数，但实际上它是另一个......
3. 被调用的匿名回调。new_callback 再次包含第二个回调函数。old_callback包含匿名函数
4. ...

现在，我想念什么？这是一些奇怪的闭合魔法还是只是一些我显然看不到的明显错误？

提前致谢!

Answer 1

为什么 old_callback 不使用 var 定义？

var old_callback = this.data_callback;

事实上，它是一个全局变量。也许它在其他地方声明过，但对我来说仍然看起来很可疑。