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Full Discussion: du
Top Forums UNIX for Dummies Questions & Answers du Post 6106 by andrec on Wednesday 29th of August 2001 07:40:38 PM
Old 08-29-2001
Question du

I have a question about the use of du. I understand that du reports the number of 512-byte blocks of disk usage used in the directory and below.

However, with only 3 small files (file a is 64 bytes; b is 68 bytes; c is 6 bytes), du seems to tell me that the OS is using 8 blocks x 512-byte to store them. Is there anything wrong with my calculations or I misunderstand anything?

When I do a 'du' with no option, it gives me:
8 .
When I do a 'du -k', it gives me:
4 .
When I do a 'du -a' , it gives me:
2 ./a
2 ./b
2 ./c
8 .



 
RK(4)							     Kernel Interfaces Manual							     RK(4)

NAME
rk - RK-11/RK03 or RK05 disk DESCRIPTION
Rk? refers to an entire disk as a single sequentially-addressed file. Its 256-word blocks are numbered 0 to 4871. Minor device numbers are drive numbers on one controller. The rk files discussed above access the disk via the system's normal buffering mechanism and may be read and written without regard to physical disk records. There is also a `raw' interface which provides for direct transmission between the disk and the user's read or write buffer. A single read or write call results in exactly one I/O operation and therefore raw I/O is considerably more efficient when many words are transmitted. The names of the raw RK files begin with rrk and end with a number which selects the same disk as the corre- sponding rk file. In raw I/O the buffer must begin on a word boundary, and counts should be a multiple of 512 bytes (a disk block). Likewise seek calls should specify a multiple of 512 bytes. FILES
/dev/rk?, /dev/rrk? BUGS
In raw I/O read and write(2) truncate file offsets to 512-byte block boundaries, and write scribbles on the tail of incomplete blocks. Thus, in programs that are likely to access raw devices, read, write and lseek(2) should always deal in 512-byte multiples. RK(4)
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