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 Integrate WunderGraph With Your Frontend Application

How to integrate WunderGraph with your frontend application

Unify and simplify APIs using WunderGraph to integrate REST, GraphQL, and databases in a single endpoint.

Boemo Mmopelwa
May 17, 2024 ⋅ 5 min read
Understanding The Latest Webkit Features In Safari 17.4

Understanding the latest Webkit features in Safari 17.4

The Safari 17.4 update brought in many modern features and bug fixes. Explore the major development-specific updates you should be aware of.

Rahul Chhodde
May 16, 2024 ⋅ 10 min read
Using Webrtc To Implement Peer To Peer Video Streaming In A Node Js Project

Using WebRTC to implement P2P video streaming

Explore one of WebRTC’s major use cases in this step-by-step tutorial: live peer-to-peer audio and video streaming between systems.

Oduah Chigozie
May 16, 2024 ⋅ 18 min read
Htmx Vs React

htmx vs. React: Choosing the right library for your project

Both htmx and React provide powerful tools for building web apps, but in different ways that are suited to different types of projects.

Temitope Oyedele
May 15, 2024 ⋅ 9 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