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Octal dump question


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 PostPosted: Tue Sep 21, 2010 9:25 pm   

Joined: Tue Sep 21, 2010 9:13 pm
Posts: 1
I have been trying to decipher something for a while now, and it eludes me...maybe you can help?

Here is the line of code:
Code:
`printf "%-24s" "${1}" | od -w99 -A n -tx1`


It is really just the "od" part, and the "-w99" that I cannot get. I believe it is formatting the ASCII string (that is what $1 contains) to an appropriate hex string, but I cannot verify.

The other question I have about this code is the "|" character...is this really doing a bitwise OR on the string, or is it a separator?

Any help would be appreciated!

JR


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 PostPosted: Tue Sep 21, 2010 11:30 pm   
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Joined: Sun Jun 27, 2010 12:57 am
Posts: 192
-w99 is an option of od to set the width of the output. In this case the output can span up to 99 characters before wrapping to a newline.

The pipe character is not a bitwise OR, but an I/O pipe between 2 shell commands. The first command, printf, outputs a string of text which then in *nix terms gets 'piped' as the input for the second command, od, which reads and converts this input. This way you can chain a lot of 'filter' commands one after the other. f.e. cat "file.txt" | grep "my string" | sed "s/my/your/" | sort


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 PostPosted: Wed Sep 22, 2010 5:50 am   

Joined: Wed Sep 22, 2010 5:35 am
Posts: 1
Thank you for that explanation patsie! It was very frustrating that I could not find specific information about that switch...

It occurred to me this morning that the "|" must be a separator within a printf statement, and that the function of the whole thing was to create a simple ASCII to hex conversion of the input. There must be a simpler way to do this, and knowing what is going on now is a step in the right direction.

Thanks again,
JR


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