2021-11-01
2092
Godson Obielum
75053
Nov 1, 2021 ⋅ 7 min read

How to protect against regex denial-of-service (ReDoS) attacks

Godson Obielum I'm a software developer with a life goal of using technology as a tool for solving problems across major industries.

Recent posts:

Why Frontend Devs Should Care About Platform Engineering

Why frontend devs should care about platform engineering

Learn how platform engineering helps frontend teams streamline workflows with Backstage, automating builds, documentation, and project management.

Muhammed Ali
Nov 3, 2025 ⋅ 6 min read
vercel ai elements featured image

How I built an AI productivity assistant with Vercel AI Elements

Build an AI assistant with Vercel AI Elements, which provides pre-built React components specifically designed for AI applications.

Emmanuel John
Nov 3, 2025 ⋅ 9 min read

How to use CSS line-clamp to trim lines of text

Master the CSS line-clamp property. Learn how to truncate text lines, ensure cross-browser compatibility, and avoid hidden UX pitfalls when designing modern web layouts.

Daniel Schwarz
Oct 30, 2025 ⋅ 3 min read
7 react Hooks you need to know

7 custom React Hooks every developer should be using

Discover seven custom React Hooks that will simplify your web development process and make you a faster, better, more efficient developer.

Murat Yüksel
Oct 30, 2025 ⋅ 8 min read
View all posts

One Reply to "How to protect against regex denial-of-service (ReDoS) attacks"

  1. Interesting article.

    Your explanation is wrong though. \w+\s* does not return “A long sentence with invalid characters that takes so much time to be matched that it potentially causes our CPU usage to increase”. it matches “A “, because \w is only a single char, so \w+ matches as many word char are available (in this case just the letter A), then \s* matches as many spaces as possible (just one in this case), the result is “A “. then (\w+\s*)* matches the whole string. It matches as many “at least one word char followed by 0 or more space”. The rest of your explanation is therefore erroneous.

    Too bad also your solution is not a real solution. It rejects rapidly the sequence with invalid chars, but it also reject any sequence with valid char ! In fact, this formula will never match anything but the empty string. This is due to the fact that you reference the 1st group from within the first group (the \1 is within the first pair of ()). If you define the first group as “The first group is the first group plus the repetition of itself”, the only solution is the empty group.

    A solution that works to you problem is “an optional blank separated list of words plus one word” and it’s spelled like this :
    /^(\w+\s+)*\w+$/
    which can be decoded as :
    ^: start
    (…)* repeat 0 or more time
    \w+: at least one word char
    \s+: at least one space char :
    \w+: followed by at least one word char
    $: then end

    It instantly matches “correct”
    it instantly matches “this is a list of word”
    it instantly does not match “this is an invalid list!”
    it instantly does not match “A long sentence with invalid characters that takes soo much time to be matched that it potentially causes our CPU usage to increase drastically!!!”

Leave a Reply

Hey there, want to help make our blog better?

Join LogRocket’s Content Advisory Board. You’ll help inform the type of content we create and get access to exclusive meetups, social accreditation, and swag.

Sign up now