I was looking at some image vulnerabilities and came across the fairly new openjpeg vulnerability in JPEG2000 images. It says that: "out of bounds memory can be accessed due to an error in mcc records parsing".

I wanted to play around with this a bit, but I cannot find anything about what mcc records are. I am assuming this is metadata information, but I cannot find what exactly I can play with here (what name it has to be and what possible values etc.). I have downloaded "JP2 Meta Editor" so I should be able to edit any metadata.

Can someone point me in the right direction about this?

EDIT: On the github page of OpenJPEG they also keep track of all the bug work and have some POC files they used to fix the problem. Using these anyone can do research on the vulnerability.

  • It helps to provide links to what you quote so that the rest of us can do the research, too.
    – schroeder
    Jan 11 '17 at 9:35

You need to read the Talos blog about the vulnerability and the source code.

mcc is an internal reference to Multiple Component Collection (line 1040).

A Multiple Component Transformation is a colour transform for images in JPEG2000.

This is not something you can edit in metadata. It's about how you craft the image itself to create an index reference error when it tries to process the colours.

Please make sure you read the CVEs for the vulnerabilities that you come across, and read the source code and bug reports so that you can understand the context for the issues.

  • Thanks for this input, I have indeed read that blog but did not understand it. With your information I can try to understand it better.
    – Wealot
    Jan 11 '17 at 10:59
  • Well interesting stuff, but it is very hard to find any (free) tools that can work with (edit) jpeg2000. And good examples are no where to be found either.... So if anyone has any suggestions? Otherwise I'll at least put this answer as the correct one as it showed correctly where to search next and answered my question. Thanks!
    – Wealot
    Jan 11 '17 at 14:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.