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 win clients with a Flutter web demo that feels real

How to win clients with a Flutter web demo that feels real

Use Flutter to build browser-based app demos that help clients visualize the product, speed up buy-in, and close deals faster.

Lewis Cianci
Mar 21, 2025 ⋅ 5 min read
TypeScript Is Getting A 10x Faster Compiler —Here’s What It Means For Developers

TypeScript is getting a 10x faster compiler — here’s what it means for developers

TypeScript’s new Go-based compiler offers 10x faster builds, improved CI/CD performance, quicker editor startup, and enhanced hot reload times.

Kapeel Kokane
Mar 21, 2025 ⋅ 6 min read

How to manage JavaScript closures in React

Learn how to manage JavaScript closures in React and explore real-life examples of proper JavaScript closure management.

Andrew Evans
Mar 21, 2025 ⋅ 10 min read
how to use css variables

How to use CSS variables like a pro

By building these four simple projects, you’ll learn how CSS variables can help you write reusable, elegant code and streamline the way you build websites.

Idorenyin Obong
Mar 20, 2025 ⋅ 18 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