JavaScript 代码技巧:foo.x 的值是什么 [英] JavaScript code trick: What's the value of foo.x

问题描述

我在GitHub前端面试题合集中发现了这个问题:

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

问题:foo.x 的值是什么?

答案是undefined.

我做了一些研究，我明白这个问题是(如果我错了，请纠正我):

• var foo = {n: 1}; 声明一个对象 foo，它的属性 n 等于 1.
• var bar = foo; 声明一个对象 bar，它引用与 foo 相同的对象.
• foo.x = foo = {n: 2}; 我相信它等于 foo.x = (foo = {n: 2});
• 然后我得到 foo.x 等于 undefined.但是，bar.x 的值是对象{n:2}.

如果bar 和foo 指向同一个对象，为什么bar.x 得到一个值而foo.x 是 未定义 吗?foo.x = foo = {n: 2}; 到底发生了什么?

解决方案

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

确定foo.x引用{n: 1}对象的属性x，赋值{n:2}到foo，并将foo的新值——{n:2}——赋给属性{n: 1} 对象的 x.

重要的是foo.x所指的foo是在foo改变之前确定的.

请参阅 ES5 规范的第 11.13.1 节:

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1. 让 lref 成为评估 LeftHandSideExpression 的结果.

2. 让 rref 成为评估 AssignmentExpression 的结果.

赋值运算符从右到左关联，所以你得到:

foo.x = (foo = {n: 2})

左侧先于右侧求值.

I found this problem in a GitHub front-end interview questions collection:

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

Question: What is the value of foo.x?

The answer is undefined.

I've done some research and what I understand this problem is (correct me if I'm wrong):

• var foo = {n: 1}; declares an object foo which has property n equal to 1.
• var bar = foo; declares an object bar which refers to the same object as foo.
• foo.x = foo = {n: 2}; which I believe is equal to foo.x = (foo = {n: 2});
• And then I got foo.x equals to undefined. However, the value of bar.x is the object {n:2}.

If bar and foo refer to same object, why did bar.x get a value while foo.x is undefined? What is really happening in foo.x = foo = {n: 2};?

解决方案

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

determines that foo.x refers to a property x of the {n: 1} object, assigns {n: 2} to foo, and assigns the new value of foo – {n: 2} – to the property x of the {n: 1} object.

The important thing is that the foo that foo.x refers to is determined before foo changes.

See section 11.13.1 of the ES5 spec:

1. Let lref be the result of evaluating LeftHandSideExpression.

2. Let rref be the result of evaluating AssignmentExpression.

The assignment operator associates right to left, so you get:

foo.x = (foo = {n: 2})

The left hand side is evaluated before the right hand side.

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