使用'let'作为变量名不会在google v8中引发任何错误 [英] Using 'let' as a variable name is not throwing any errors in google v8

问题描述

我在Chrome开发者控制台中写了一些随机代码。令我惊讶的是，chrome允许我使用 let 作为变量名，这是完全错误的，因为 let 是一个保留关键字。我需要了解为什么会发生这种情况。

I was writing some random code in the chrome developer console. For my surprise, chrome let me use let as a variable name which is completely wrong as let is a reserved keyword. I need to understand why is this happening.

方案：

Scenarios:

var const = 78 //throws an error as expected

var function = 46 //throws an error as expected

var let = 56 //didn't throw an error :O

let //prints 56, which is wrong because 'let' is a keyword

let ab = 90

ab //prints 90 as expected

这个缺陷存在于节点 。但是，当我在 Babel REPL 这是一个错误。

This flaw exists in node. But, when I try it in Babel REPL it is throwing an error.

我认为这与Google v8有关

I think this is something to do with Google v8

推荐答案

一个很好的背后推理可以在 Mohsen Azimi的这篇文章。下面是关于它的简要总结。

A nice write-up of the reasoning behind this can be found in this article by Mohsen Azimi. Here's a quick summary of it.

以下关键字在JavaScript规范中定义为 FutureReservedWord ：

The following keywords are defined in the JavaScript spec as FutureReservedWord:

implements     interface   let       package    private
protected      public      static    yield

在正常模式下，这些可以用作无错误的变量名称;但是，在严格模式下，它们被视为保留字，并会引发以下错误：

In normal mode, these can be used as variable names without errors; however, in strict mode they are treated as reserved words and will throw the following error:

SyntaxError: Cannot use the reserved word 'let' as a variable name in strict mode.

这样ES2015之前的代码不会中断 - 如果某人已经命名了很多变量让在传统应用程序中，如果JS规范突然打破了所有内容，他们可能不会开心。

This is so that pre-ES2015 code doesn't break - if someone had named lots of their variables let in a legacy app, they probably wouldn't be happy if the JS spec suddenly broke everything.

这篇关于使用'let'作为变量名不会在google v8中引发任何错误的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！