Question

我有一个原型，有一个方法可以添加回调：

/*
 * 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，我正在添加三个回调函数。但是，如果最终为元素调用this.data_callback（），则会导致整个事件无限循环。在我尝试跟踪错误时，上面的警报（是的，我知道有调试工具，但这样更舒服）提供了以下对此问题的见解：

调用匿名函数/闭包（包含jQuery.extend（）的函数）。 new_callback包含第三个回调函数。 old_callback包含另一个......
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，但实际上它还是另一个...
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
...

现在，我想念什么？这是一些奇怪的封闭魔法还是一些我显然无法看到的明显错误？

提前致谢！

Answer 1

为什么old_callback未定义var？

var old_callback = this.data_callback;

实际上，它是一个全局变量。也许它在其他地方宣布，但它仍然看起来很可疑。

奇怪的递归无限循环我无法追踪

1 个答案: