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 fix React routing loopholes with the React Router Middleware

How to fix React routing loopholes with the React Router Middleware

Learn how React Router’s Middleware API fixes leaky redirects and redundant data fetching in protected routes.

Ikeh Akinyemi
Nov 13, 2025 ⋅ 3 min read
How I used Mastra to build a prize-winning RAG agent

How I used Mastra to build a prize-winning RAG agent

A developer’s retrospective on creating an AI video transcription agent with Mastra, an open-source TypeScript framework for building AI agents.

Chinwike Maduabuchi
Nov 13, 2025 ⋅ 12 min read

Ensuring frontend data integrity with TanStack DB transactions

Learn how TanStack DB transactions ensure data consistency on the frontend with atomic updates, rollbacks, and optimistic UI in a simple order manager app.

Emmanuel John
Nov 13, 2025 ⋅ 11 min read
the replay november 12

The Replay (11/12/25): Stop making these useEffect mistakes

Discover what’s new in The Replay, LogRocket’s newsletter for dev and engineering leaders, in the November 5th issue.

Matt MacCormack
Nov 12, 2025 ⋅ 33 sec 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