JavaScript:使用没有运算符“new"的构造函数 [英] JavaScript: using constructor without operator 'new'

问题描述

请帮助我理解为什么以下代码有效:

Please help me to understand why the following code works:

<script>
    var re = RegExp('\ba\b') ;
    alert(re.test('a')) ;
    alert(re.test('ab')) ;
</script>

第一行没有 new 操作符.

In the first line there is no new operator.

据我所知，JavaScript 中的构造函数是一个函数，它初始化由运算符 new 并且它们不打算返回任何内容.

As far as I know, a contructor in JavaScript is a function that initialize objects created by the operator new and they are not meant to return anything.

推荐答案

一般来说，如果某物被记录为构造函数，则将 new 与它一起使用.但在这种情况下，RegExp 为您将其作为函数调用的情况定义了工厂"行为.请参阅 ECMAScript (JavaScript) 规范(链接到即将发布的规范；新规范中的部分编号相同，您可以从 ECMA front 下载页面 [在右侧]；我不想直接链接到大约 4MB 的 PDF 文件):

In general, if something is documented as being a constructor, use new with it. But in this case, RegExp has a defined "factory" behavior for the situation where you've called it as a function instead. See Section 15.10.3 of the ECMAScript (JavaScript) specification (that links to the outgoing spec; the section number is the same in the new spec, which you can download from the ECMA front page [on the right-hand side]; I don't want to directly link to a ~4MB PDF file):

15.10.3 作为函数调用的 RegExp 构造函数
15.10.3.1 RegExp(pattern, flags)
如果 pattern 是一个对象 R，其 [[Class]] 属性为 "RegExp" 并且标志未定义，则返回 R 不变.否则调用 RegExp 构造函数 (15.10.4.1)，将模式和标志参数传递给它，并返回由该构造函数构造的对象.

15.10.3 The RegExp Constructor Called as a Function
15.10.3.1 RegExp(pattern, flags)
If pattern is an object R whose [[Class]] property is "RegExp" and flags is undefined, then return R unchanged. Otherwise call the RegExp constructor (15.10.4.1), passing it the pattern and flags arguments and return the object constructed by that constructor.

您实际上可以定义自己的 JavaScript 构造函数以允许省略 new 关键字(通过检测它们已被作为函数调用，然后正确调用自己)，但我不会不要建议它，因为它会导致误导性代码.

You can actually define your own JavaScript constructor functions to allow omitting the new keyword (by detecting they've been called as a function instead, and turning around and calling themselves correctly), but I wouldn't suggest it as it leads to misleading code.

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