Question

我没有完全理解这一点：

var one = {};
var two = {};
var three = {};

three[one] = "one";
three[two] = "two";

console.log(three[one]); // Alerts "two"

据我了解，JS需要在括号属性表示法中使用一个字符串。所以看起来JS正试图改变：

three[one] = "one";
three[two] = "two";

成：

three[one.toString()] = "one";
three[two.toString()] = "two";

两种toString（）方法都没有定义，所以它们最终是相同的值？不确定我是否明白了。

Answer 1

实际上定义了toString方法，并为两者（[object Object]）生成相同的字符串。

＆＃13;

var one = {};
var two = {};
$('body').append(one.toString()+'<br>'+two.toString());

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

＆＃13;

Answer 2

var one = {};
var two = {};
var three = {};

three[one] = "one";
three[two] = "two";

最后两行说：

three[object] = "one";
three[object] = "two";

因为一个和两个都被声明为对象。

您正在制作的三个对象将如下所示：

[object Object] {
  [object Object]: "two"
}