为什么在foo.x = foo = {n：2}中未定义foo.x的值？

问题描述

此代码：

var foo = {n: 1};
var bar = foo;
foo.x = foo = {n: 2};

您能否解释一下其含义：

Can you please explain what is meant by:

foo.x = foo = {n: 2};

我看到 {n：2} 被分配到 foo 。为什么 undefined 分配给 foo.x ？ foo = {n：2}; 返回 undefined ？

I see that {n:2} is assigned to foo. Why is undefined assigned to foo.x? Does foo = {n: 2}; return undefined?

推荐答案

，即使操作员具有从右到左的优先级，也会首先评估赋值表达式的左侧。因此表达式 foo.x = foo = {n：2} 与 foo.x =（foo = {n：2}相同）的评估如下：

According to the spec, the left hand side of an assignment expression is evaluated first, even though the operator has right-to-left precedence. Thus the expression foo.x = foo = {n: 2} which is the same as foo.x = (foo = {n: 2}) is evaluated like this:

1. 评估左手表达式 foo.x 获取引用，这是右手表达式的值将被分配到的位置。
   
   


2. 评估权利-hand expression，以获取将要分配的值。右侧是另一个赋值表达式，因此它的评估方式相同：
   
   
   
   

1. Evaluate the left-hand expression foo.x to get a reference, which is where the value of the right-hand expression will be assigned to.
   
   
2. Evaluate the right-hand expression, to to get the value that will be assigned. The right-hand side is another assignment expression, so it gets evaluated the same way:
   
   

1. 评估 foo 确定分配到何处。


2. 评估表达式 {n：2} ，其中创建一个对象，以确定要分配的值。


3. 将 {n：2} 分配给foo，并返回 {n：2} 。
   
   

1. Evaluate foo to determine where to assign to.
2. Evaluate the expression {n:2}, which creates an object, to determine the value to assign.
3. Assign {n:2} to foo, and return {n:2}.
   
   

分配值右侧的表达式评估为（ {n：2} ），引用为 foo.x 已解决到步骤1（ foo 分配了新值）。这与 bar.x 也是一样的，因为之前的行上的赋值 bar = foo 。

Assign the value that the expression on the right-side evaluated to ({n:2}), to the reference that foo.x resolved to in step 1 (before foo was assigned a new value). Which is also the same as bar.x, because of the assignment bar = foo on the line before.

完成后，原始对象 bar 仍然是对，将有一个 x 属性引用创建的第二个对象。 foo 也是对第二个对象的引用，因此 foo === bar.x 。

When this is done, the original object, that bar is still a reference to, will have an x property that references the second object created. foo is also a reference to that second object, so foo === bar.x.

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