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Old 05-05-2009, 02:33 AM   #1
schneidz
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deciphering dd output


hi, i have a bunch of output like this after running this command:

Code:
[schneidz@hyper schneidz]$ ll fc9-live-usb.iso.bz2
-rw-rw-r--    1 schneidz schneidz 2119848361 Mar  2 21:34 fc9-live-usb.iso.bz2

[schneidz@hyper schneidz]$ bunzip2 -c fc9-live-usb.iso.bz2 | dd bs=16K of=/dev/sdb & pid=$!
[schneidz@hyper schneidz]$ while [ 1 ]
> do
> kill -USR1 $pid
> sleep 5
> done

210+719388 records out
210+725564 records in
210+725563 records out
210+731282 records in
210+731281 records out
1335+733472 records in
1335+733471 records out
3526+734906 records in
3526+734906 records out
4514+737367 records in
4514+737366 records out
4560+741806 records in
4560+741806 records out
4560+749750 records in
4560+749749 records out
4560+756515 records in
4560+756515 records out
4560+763826 records in
4560+763826 records out
4560+771637 records in
4560+771636 records out
4812+774508 records in
4812+774508 records out
what does all this stuff mean ?
thx
 
Old 05-05-2009, 04:31 AM   #2
JulianTosh
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Looks like you're writing a bootable disk image to a usb flash drive. The "records in/out" is just dd's output when it is interupted - telling you how much data was read/written.
 
Old 05-05-2009, 11:46 AM   #3
schneidz
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^ thx, but does this number indicate number of bits, number of bytes, number of blocks, ...

why would they write it like 210+719388 (does this mean that 719598 bits have been written so far ?)

is 210 some starting memory address and 719388 is the offset from that address ?

this info is not in the man page.

Last edited by schneidz; 05-05-2009 at 11:47 AM.
 
Old 05-05-2009, 10:07 PM   #4
JulianTosh
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Quote:
DIAGNOSTICS

Upon completion, dd reports the number of input and output records:

f+p records in

Number of full and partial blocks read.
f+p records out

Number of full and partial blocks written.

When conv=block is specified and there is at least one truncated block, the number of truncated records is also reported:

n truncated records
Block size in this case was 16k.
 
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Old 05-06-2009, 03:21 PM   #5
schneidz
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almost there.
in the above example the total size is 2119848361 (2 gigabytes) and each block is 16k.

should that mean i can predict that the total amount of records would be 2 gb / 16 kb = 131072

why does it indicate its already past 700,000 records ?
 
Old 05-06-2009, 05:28 PM   #6
JulianTosh
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The file (*.bz2) is a compressed version of the disk image (think zip archive). The amount of data written to the disk is going to be greater than then amount of data in the .bz2 file.
 
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Old 05-07-2009, 12:34 AM   #7
schneidz
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^ oh yeah, good catch.
 
Old 06-03-2009, 01:39 PM   #8
schneidz
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this is interesting:
Code:
[schneidz@hyper schneidz]$ sudo dd if=/dev/mem | wc -c
1047456+0 records in
1047456+0 records out
536297472
[schneidz@hyper schneidz]$ expr 1047456 \* 512
536297472
[schneidz@hyper schneidz]$
 
  


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