Number of interrupts last second (Bash)

Question

I'm programming in Bash and I'm trying to find out how many interrupts that happend the last second.

I've searched and tried to find how many interrupts total, so I could use sleep 1 and calculate the difference, but with no success.

Thanks for your help.

Answer 1

This answer is similar to your other question. Once again, vmstat(8) will be our friend.

Example output:

$ vmstat 1 2
procs -----------memory---------- ---swap-- -----io---- -system-- ----cpu----
 r  b   swpd   free   buff  cache   si   so    bi    bo   in   cs us sy id wa
 0  0    752 135436  31276 392252    0    0     2     1   40  101  0  0 99  0
 0  0    752 138404  31396 392816    0    0     0     0   68  188  1  1 98  0

The first line is the average since reboot, the second line is a sampling of the last second.

From the manpage:

   System
       in: The number of interrupts per second, including the clock.
       cs: The number of context switches per second.

We can use awk(1) to parse the output:

ints=$(vmstat 1 2 | tail -1 | awk '{print $11}')

This should work on Linux, FreeBSD, and probably MacOSX.

Answer 2

Just a food for thought you can do something like this

watch "cat /proc/interrupts |awk '{print \$1,\$2}'"

By passing -n to watch command you dont need sleep option and you can have decide when you get an output

-n, --interval