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:

the replay december 10

The Replay (12/10/25): Fixing AI code, over-engineering JavaScript, and more

Fixing AI code, over-engineering JavaScript, and more: discover what’s new in The Replay, LogRocket’s newsletter for dev and engineering leaders, in the December 10th issue.

Matt MacCormack
Dec 10, 2025 ⋅ 33 sec read

How to use TOON to reduce your token usage by 60%

TOON is a lightweight format designed to reduce token usage in LLM prompts. This post breaks down how it compares to JSON, where the savings come from, and when it actually helps.

Rosario De Chiara
Dec 10, 2025 ⋅ 5 min read
Fixing AI Generated Code

Fixing AI-generated code: 5 ways to debug, test, and ship safely

Andrew Evans, principal engineer and tech lead at CarMax discusses five ways to fix AI-generated code and help you debug, test, and ship safely.

Andrew Evans
Dec 10, 2025 ⋅ 9 min read
Apple Liquid Glass LogRocket

How to create Liquid Glass effects with CSS and SVG

This tutorial walks through recreating Apple’s Liquid Glass UI on the web using SVG filters, CSS, and React. You’ll learn how to build refraction and reflection effects with custom displacement and specular maps, and how to balance performance and accessibility when using advanced filter pipelines.

Rahul Chhodde
Dec 8, 2025 ⋅ 10 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