console.log 中的错误?

Bug in console.log?(console.log 中的错误?)
本文介绍了console.log 中的错误?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

可能重复:
Chrome 的 JavaScript 控制台是否懒于评估数组?

我尝试以下代码:

var myList = new Object();
var item   = new Object();
item.text  = "item-1";
myList[3]  = item;

console.log(myList);
console.log(myList[3].text);

// Assign another object to the same entry
var item2   = new Object();
item2.text  = "item-2";
myList[3]  = item2;

console.log(myList);
console.log(myList[3].text);

结果很奇怪:

* Object
  * 3: Object
      text: "item-2"

item-1

* Object
  * 3: Object
      text: "item-2"

item-2

但是 - 如果我在一段时间后执行第二部分(使用 setTimeout)并展开第一个对象,我做对了,即:

BUT - if i execute the second part after some time (using setTimeout), and unfold the first object, I get it right, i.e.:

* Object
  * 3: Object
      text: "item-1"

item-1

* Object
  * 3: Object
      text: "item-2"

item-2

我觉得分享它很重要,因为我认为人们可能会浪费大量时间来尝试理解他的代码中的问题.如果有人提到了一个开放的错误或其他东西 - 请回复这张票.谢谢!

I find it important to share it, since I think one can waste a lot of time trying to understand what's wrong in his code. And if somebody has some reference to an open bug or something - please reply this ticket. Thanks!

推荐答案

我的观点是,这是一个非常烦人的功能",我真的希望我可以关闭它,它使调试成为一场噩梦,不知道在什么时候时间可能已经更新了一个对象,同时试图在代码中的给定点建立精确的对象状态.该功能可能对观察点"等有用,但不适用于所谓的日志"(线索就在名称中).

My view is that this is a horrendously irritating 'feature' that I really wish I could turn off, it makes debugging a nightmare, not knowing at which point in time something may have updated an object, whilst trying to establish exact object state at a give point in the code. The feature could be useful for 'watch points' etc, but not in something called a 'LOG' (the clue is in the name).

考虑这个代码片段:

var person = {'name':'Tom'};
console.log( person);  //output the entire object variable
person.name = 'Thomas';
//the output is an object, whose 'name' value is 'Thomas', even though the log statement was placed before the value was changed to 'Thomas'.

然后:

var person = {'name':'Tom'};
console.log( person.name);    //changed to output a string variable
person.name = 'Thomas';
//the output here, however, has not dynamically updated and correctly outputs 'Tom'

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