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 HTML’s Selectedcontent Element Improves Dropdowns

​​How HTML’s <selectedcontent> element improves dropdowns

is an experimental HTML element that gives developers control over how a selected option is displayed, using just HTML and CSS.

Temitope Oyedele
Jun 27, 2025 ⋅ 6 min read
advanced caching in Node.js with Valkey

How to get faster data access in Node.js with Valkey

Learn how to implement an advanced caching layer in a Node.js app using Valkey, a high-performance, Redis-compatible in-memory datastore.

Muhammed Ali
Jun 27, 2025 ⋅ 7 min read
how to properly handle rejected promises in TypeScript

How to properly handle rejected promises in TypeScript

Learn how to properly handle rejected promises in TypeScript using Angular, with tips for retry logic, typed results, and avoiding unhandled exceptions.

Lewis Cianci
Jun 26, 2025 ⋅ 4 min read
Your AI Has Agency — Here’s How To Architect Its Frontend

Your AI has agency — here’s how to architect its frontend

AI’s not just following orders anymore. If you’re building the frontend, here’s how to design interfaces that actually understand your agent’s smarts.

Rosario De Chiara
Jun 25, 2025 ⋅ 5 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