[arch-general] most efficient way to get linux kernel statistics

[Andre Schmidt]

hello archers,

thought i ask here first, before i try some kernel mailing list.

i'm writing (for fun) a tiny daemon that sends linux usage (cpu, mem, net, etc.) statistics as efficiently as possible to another machine, repeatedly, to get "live" data.

at the moment i'm simply sending /proc files, but they sometimes have too much data. so was wondering if there is a more efficient way to get (only parts of) the data thats available in /proc?

for example, /proc/meminfo has all this info:

MemTotal:        8051660 kB
MemFree:         6046020 kB
MemAvailable:    6843080 kB
Buffers:          109116 kB
Cached:           708336 kB
SwapCached:            0 kB
Active:          1332416 kB
Inactive:         455428 kB
Active(anon):     971040 kB
Inactive(anon):    22340 kB
Active(file):     361376 kB
Inactive(file):   433088 kB
Unevictable:          16 kB
Mlocked:              16 kB
SwapTotal:       1048572 kB
SwapFree:        1048572 kB
Dirty:                16 kB
Writeback:             0 kB
AnonPages:        970412 kB
Mapped:           323384 kB
Shmem:             23008 kB
Slab:              71760 kB
SReclaimable:      44860 kB
SUnreclaim:        26900 kB
KernelStack:        6544 kB
PageTables:        22924 kB
NFS_Unstable:          0 kB
Bounce:                0 kB
WritebackTmp:          0 kB
CommitLimit:     5074400 kB
Committed_AS:    3487604 kB
VmallocTotal:   34359738367 kB
VmallocUsed:           0 kB
VmallocChunk:          0 kB
HardwareCorrupted:     0 kB
AnonHugePages:    413696 kB
HugePages_Total:       0
HugePages_Free:        0
HugePages_Rsvd:        0
HugePages_Surp:        0
Hugepagesize:       2048 kB
DirectMap4k:      168600 kB
DirectMap2M:     8091648 kB

but i only need:

MemTotal:        8051660 kB
MemAvailable:    6843080 kB
Cached:           708336 kB

event better would be if kernel would only tell me:

8051660 6843080 708336

if /proc is the only way to get this info, i wonder if creating a kernel module for this would be more efficient, or even possible?

Cheers
Andre Schmidt