CODEWITHSUNDEEP
linuxbashgnu-coreutils
David Portabella
David Portabella

Reputation: 12710

get a part of a binary file using gnu-coreutils, bash

I want to get a part of a binary file, from byte #480161397 to #480170447 (included, 9051 bytes in total)

I use cut -b, and I expected the size of trunk1.gz to be 9051 bytes, but I get a different result.

$ wget https://commoncrawl.s3.amazonaws.com/crawl-data/CC-MAIN-2016-07/segments/1454701152097.59/warc/CC-MAIN-20160205193912-00264-ip-10-236-182-209.ec2.internal.warc.gz

$ cut -b480161397-480170447 CC-MAIN-20160205193912-00264-ip-10-236-182-209.ec2.internal.warc.gz >trunk1.gz

$ echo $((480170447-480161397+1))
9051

$ ls -l trunk1.gz
-rw-r--r--  1 david  staff     3400324 Sep  8 10:28 trunk1.gz

What is wrong?

Upvotes: 2

Views: 261

Answers (2)

oliv
oliv

Reputation: 13249

If you work with binary, I advise you to use dd command.

dd if=trunk1.gz bs=1 skip=480161397 count=9051 of=output.bin

bs is the block size and is set to 1 byte.

Upvotes: 1

Leon
Leon

Reputation: 32474

cut -bN-M copies the range N-M bytes from every line of the input.

Example:

$ cut -b4-7 <<END
0123456789
abcdefghij
ABCDEFGHIJ
END

Output:

3456
defg
DEFG

Consider using dd for your purposes.

Upvotes: 2

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