Why do regex constructors need to be double escaped?

Question

In the regex below, \s denotes a space character. I imagine the regex parser, is going through the string and sees \ and knows that the next character is special.

But this is not the case as double escapes are required.

Why is this?

var res = new RegExp('(\\s|^)' + foo).test(moo);

Is there a concrete example of how a single escape could be mis-interpreted as something else?

Answer 1

You are constructing the regular expression by passing a string to the RegExp constructor.

\ is an escape character in string literals.

The \ is consumed by the string literal parsing…

const foo = "foo";
const string = '(\s|^)' + foo;
console.log(string);

… so the data you pass to the RegEx compiler is a plain s and not \s.

You need to escape the \ to express the \ as data instead of being an escape character itself.

Answer 2

Inside the code where you're creating a string, the backslash is a javascript escape character first, which means the escape sequences like \t, \n, \", etc. will be translated into their javascript counterpart (tab, newline, quote, etc.), and that will be made a part of the string. Double-backslash represents a single backslash in the actual string itself, so if you want a backslash in the string, you escape that first.

So when you generate a string by saying var someString = '(\\s|^)', what you're really doing is creating an actual string with the value (\s|^).

Answer 3

The Regex needs a string representation of \s, which in JavaScript can be produced using the literal "\\s".

Here's a live example to illustrate why "\s" is not enough:

alert("One backslash:          \s\nDouble backslashes: \\s");

Note how an extra \ before \s changes the output.

Answer 4

\ is used in Strings to escape special characters. If you want a backslash in your string (e.g. for the \ in \s) you have to escape it via a backslash. So \ becomes \\ .

EDIT: Even had to do it here, because \\ in my answer turned to \.