In article <[EMAIL PROTECTED]> Andrew
Hood <[EMAIL PROTECTED]> writes:
>On Sat, 10 Nov 2007 00:38:23 +0000, Per Hedeland wrote:
>
>> In article <[EMAIL PROTECTED]>
>> [EMAIL PROTECTED] (Hal Murray) writes:
>>>
>>>>I known now, in the "recen"t kernel the internal frequency will be to
>>>>250 Mhz...
>>>>With my gentoo it's not a problem for me because i make myself my kernel
>>>>... but if i take Mandriva or a another distribution how find this
>>>>values ?
>>>
>>>I assume you mean the scheduling clock which would be 250 Hz rather than
>>>MHz.
>>>
>>>One way is to look in the kernel souces.
>>>
>>>Another way is to measure it. Write a program that usleeps for a
>>>microsecond and see how long it really takes. (Do it in a loop for 1000
>>>tries and print out a histogram.)
>>
>> On Linux, a simpler way can be to look at /proc/interrupts - e.g.
>> (probably Linux-version- and possibly config-specific):
>>
>> $ (cat /proc/interrupts; sleep 10; cat /proc/interrupts) | \
>> awk '/timer/{prev=now; now=$2} END{printf "%dHz\n", int((now-prev)/10)}'
>
>This doesn't work on multi-CPU systems.
Actually the system I verified it on (mentioned elsewhere in the thread)
was dual-CPU...
>: cat /proc/interrupts |grep timer
> 0: 129602418 129428516 IO-APIC-edge timer
...but you're right of course - it just so happened that all the "timer"
interrupts were fielded on CPU 0 on the system where I tried it:
$ cat /proc/interrupts |grep timer
0: 3845082271 0 IO-APIC-edge timer
(Probably not a random thing, but I don't know what causes it - there's
certainly no "end user" config to make it so.)
>You need to sum across all CPUs. A bit more complex:
>
>-------------- snip --------------
>
>#!/usr/bin/perl -w
It's still possible to do with a one-liner awk, just that the line gets
a tad longer.:-) Left as an exercise etc...
--Per Hedeland
[EMAIL PROTECTED]
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