使用watch.js解开对象的问题

Question

我正试图在下面的代码中使用watch.js。这个想法也是以一种方式运行watch.js，因此它检测到（vars.in_animation的）布尔变化，然后解开并退出。但是，会发生什么情况是代码陷入了无限循环，每次vars.in_animation都会触发监视回调，并且监视代码永远不会运行。

我不知道问题是我使用watch.js的方式，还是在代码结构中存在其他问题。有任何想法吗？

base.goTo = function(targetSlide){ 
    if (vars.in_animation || !api.options.slideshow) 
    {    
    //arrange for a callback once in_animation is false again 
    //only need to detect first change from true to false, and then unwatch 
    watch(vars, "in_animation", function() 
     { 
     console.log('called back, value of in_animation ' + vars.in_animation); 
     unwatch(vars,"in_animation",vars.alert); //never called 
     base.goTo2(vars.saved_slide); 
     }); 

     vars.saved_slide = targetSlide; 
     return false; 
    }  
    else { base.goTo2(targetSlide); }    
    } 

编辑，刚才提出的问题的小提琴>http://jsfiddle.net/SZ2Ut/3/

var obj = { 
name: "buddy", 
alert: function(){ 
    alert(obj.phrase + " " + obj.name); 
} 

} 

watch(obj, "name", function(){ 
alert('name has changed to: ' + obj.name); 
}); 

//should only fire once for this one 
obj.name = "johnny"; 
unwatch(obj, "name", obj.alert); 
obj.name = "phil"; 

Answer 1

unwatch似乎期望应作为其第三个参数，可以删除观察者回调。尝试

watch(vars, "in_animation", function callback() { 
    console.log('called back, value of in_animation ' + vars.in_animation); 
    unwatch(vars,"in_animation", callback); 
    base.goTo2(vars.saved_slide); 
});