perl - binary sed replacement

Question

I was attempting to do a sed replacement in a binary file however I am beginning to believe that is not possible. Essentially what I wanted to do was similar to the following:

sed -bi "s/(xFFxD8[[:xdigit:]]{1,}xFFxD9)/1/" file.jpg

The logic I wish to achieve is: scan through a binary file until the hex code FFD8, continue reading until FFD9, and only save what was between them (discards the junk before and after, but include FFD8 and FFD9 as the saved part of the file)

Is there a good way to do this? Even if not using sed?

EDIT: I just was playing around and found the cleanest way to do it IMO. I am aware that this grep statement will act greedy.

hexdump -ve '1/1 "%.2x"' dirty.jpg | grep -o "ffd8.*ffd9" | xxd -r -p > clean.jpg

Answer 1

bbe is a "sed for binary files", and should work more efficiently for large binary files than hexdumping/reconstructing.

An example of its use:

$ bbe -e 's/original/replaced/' infile > outfile

Further information on the man page.