Question about temp partition and swap files .

[microsky]

Under Solaris 2.6,I found a problem when I calculate the
swap files size ,as following showing ,in the feedback of
command 'swap -l' the swap file size is
3669984*512=1834992(Kbytes) .But in the command 'swap -s'
it show me the swap space is
1621176k+2737240k=4358416(Kbytes) .In command of 'df -k' ,
it show me the location of swap partition is
/dev/md/dsk/d1 and the size is 1.75Gbytes .
Anybody know which one is correct,why ?
Thanks !
=======================================================
$ swap -l
swapfile dev swaplo blocks free
/dev/md/dsk/d1 85,1 16 3669984 2936400
$ swap -s
total: 1572336k bytes allocated + 48840k reserved = 1621176k used, 2737240k available
=======================================================

 

[bluedevils]

The man page will tell you swap -s includes physical memory and -l does not. If you're looking for swap file system, then I would guess to use swap -l.

 

[microsky]

Thank you very much!

 

[FarahRegal]

Reading the output of the swap -s command is often confusing because the command mixes up the terms reserved and allocated
swap -s commnads reads the kernel directly to obtain a snapshot of current anoninfo values

read more this article

http://www.itworld.com/Comp/2377/UIR980701perf/

.

Farah regal
good luck
"think twice and hit enter once"

 

[marrow]

You can also run vmstat: -

vmstat 2 5

will run command 5 times at 2 second intervals and display swap and free memory in
4yth & 5th column. Shows amount of swap space currently available. Please ignore the 1st line of data displayed though.