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 开发者控制台中编写了一些随机代码.令我惊讶的是，chrome 让我使用 let 作为变量名，这是完全错误的，因为 let 是保留关键字.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.

场景:

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node.但是，当我在 Babel REPL 抛出错误.

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

可以在 这篇文章来自 Mohsen Azimi.这是它的快速摘要.

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

在普通模式下，这些可以用作变量名而不会出错；但是，在严格模式下，它们被视为保留字，并会抛出以下错误:

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.

这样做是为了使 ES2015 之前的代码不会被破坏——如果有人在遗留应用程序中命名了很多变量 let，如果 JS 规范突然被破坏，他们可能会不高兴一切.

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.

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