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:

How to speed up long lists with TanStack Virtual

How to speed up long lists with TanStack Virtual

Build fast, scalable UIs with TanStack Virtual: virtualize long lists, support dynamic row heights, and implement infinite scrolling with React.

Ikeh Akinyemi
Nov 28, 2025 ⋅ 8 min read
why you should ci cd your project from day one

Why you should set up CI/CD from day one for your apps

CI/CD isn’t optional anymore. Discover how automated builds and deployments prevent costly mistakes, speed up releases, and keep your software stable.

Lewis Cianci
Nov 28, 2025 ⋅ 9 min read

Top 5 AI code review tools in 2025

A quick comparison of five AI code review tools tested on the same codebase to see which ones truly catch bugs and surface real issues.

Emmanuel John
Nov 27, 2025 ⋅ 7 min read
css corner shape property

How to create fancy corners using CSS corner-shape

Learn about CSS’s corner-shape property and how to use it, as well as the more advanced side of border-radius and why it’s crucial to using corner-shape effectively.

Daniel Schwarz
Nov 26, 2025 ⋅ 7 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

Would you be interested in joining LogRocket's developer community?

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