问题描述
我对变量分配有疑问
var foo = {n: 1};
foo.x = foo = {n: 2};
console.log(foo.x) // undefined
为什么foo.x是undefined?
推荐答案
=赋值运算符是右关联的，这意味着赋值链是从右到左求值的.也就是说，语言会这样对待
The = assignment operator is right-associative, meaning that a chain of assignments is evaluated right to left. That is, the language treats this
x = y = x+2;
好像是写的
x = (y = x+2);
因此，首先为y分配加法结果(5)，然后为x分配该赋值的值，该值也为5.
So first y is assigned the result of the addition (5), and then x is assigned the value of that assignment, which is also 5.
问题中的第一个例子是同一个故事，但事情要复杂一些:
The first example in your question is the same story, but things are a little more complicated:
foo.x = foo = { n: 2 };
首先进行的是对象文字的foo分配.但是，在此之前， 语言将确定foo.x的引用值，该引用值是对原始上的(尚不存在)属性的引用. c7>对象.因此，当对该属性进行右手赋值时，它确实起作用，但是该属性是在旧对象上设置的，而不是在新对象上设置的！
The first assignment that's made is to foo, of that object literal. However, before that point, the language will have determined the reference value of foo.x, which is a reference to a (not-yet existing) property on the original foo object. Thus, when the value of the right-hand assignment is made to that property, it does work, but the property is set on the old object, not the new one!
让我们更改示例:
var foo = {n: 1}, foo2 = foo;
foo.x = foo = {n: 2};
console.log(foo2.x); // { n: 2 }
该版本保留了对变量foo2中foo原始值的另一个引用.即使foo在双重分配中被覆盖，foo2仍将继续引用原始的foo.
That version preserves another reference to the original value of foo in the variable foo2. Even though foo is overwritten in the double assignment, foo2 will continue to refer to the original foo.
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