How do I manually convert jiffies to milliseconds and vice versa in Linux? I know kernel 2.6 has a function for this, but I'm working on 2.4 (homework) and though I looked at the code it uses lots of macro constants which I have no idea if they're defined in 2.4.
As a previous answer said, the rate at which jiffies increments is fixed.
The standard way of specifying time for a function that accepts jiffies is using the constant HZ.
That's the abbreviation for Hertz, or the number of ticks per second. On a system with a timer tick set to 1ms, HZ=1000. Some distributions or architectures may use another number (100 used to be common).
The standard way of specifying a jiffies count for a function is using HZ, like this:
schedule_timeout(HZ / 10); /* Timeout after 1/10 second */
In most simple cases, this works fine.
2*HZ /* 2 seconds in jiffies */
HZ /* 1 second in jiffies */
foo * HZ /* foo seconds in jiffies */
HZ/10 /* 100 milliseconds in jiffies */
HZ/100 /* 10 milliseconds in jiffies */
bar*HZ/1000 /* bar milliseconds in jiffies */
Those last two have a bit of a problem, however, as on a system with a 10 ms timer tick, HZ/100 is 1, and the precision starts to suffer. You may get a delay anywhere between 0.0001 and 1.999 timer ticks (0-2 ms, essentially). If you tried to use HZ/200 on a 10ms tick system, the integer division gives you 0 jiffies!
So the rule of thumb is, be very careful using HZ for tiny values (those approaching 1 jiffie).
To convert the other way, you would use:
jiffies / HZ /* jiffies to seconds */
jiffies * 1000 / HZ /* jiffies to milliseconds */
You shouldn't expect anything better than millisecond precision.
Jiffies are hard-coded in Linux 2.4. Check the definition of HZ, which is defined in the architecture-specific param.h. It's often 100 Hz, which is one tick every (1 sec/100 ticks * 1000 ms/sec) 10 ms.
This holds true for i386, and HZ is defined in include/asm-i386/param.h.
There are functions in include/linux/time.h called timespec_to_jiffies and jiffies_to_timespec where you can convert back and forth between a struct timespec and jiffies:
#define MAX_JIFFY_OFFSET ((~0UL >> 1)-1)
static __inline__ unsigned long
timespec_to_jiffies(struct timespec *value)
{
unsigned long sec = value->tv_sec;
long nsec = value->tv_nsec;
if (sec >= (MAX_JIFFY_OFFSET / HZ))
return MAX_JIFFY_OFFSET;
nsec += 1000000000L / HZ - 1;
nsec /= 1000000000L / HZ;
return HZ * sec + nsec;
}
static __inline__ void
jiffies_to_timespec(unsigned long jiffies, struct timespec *value)
{
value->tv_nsec = (jiffies % HZ) * (1000000000L / HZ);
value->tv_sec = jiffies / HZ;
}
Note: I checked this info in version 2.4.22.
I found this sample code on kernelnewbies. Make sure you link with -lrt
#include <unistd.h>
#include <time.h>
#include <stdio.h>
int main()
{
struct timespec res;
double resolution;
printf("UserHZ %ld\n", sysconf(_SC_CLK_TCK));
clock_getres(CLOCK_REALTIME, &res);
resolution = res.tv_sec + (((double)res.tv_nsec)/1.0e9);
printf("SystemHZ %ld\n", (unsigned long)(1/resolution + 0.5));
return 0;
}
To obtain the USER_HZ value (see the comments under the accepted answer) using CLI:
getconf CLK_TCK
Related
I am trying to setup a periodic timer triggering a function every seconds, but there is a small drift between each call. After some investigations, I found that this is the add_timer() call which adds an offset of 2 to the expires field (~2ms in my case).
Why is this drift added? Is there a clean way to prevent it? I am not trying to get an accurate millisecond precision, I have a vague understanding of the kernel real-time limitations, but at least to avoid this intentional delay at each call.
Here is the output from a test module. Each couple of numbers is the value of the expires field just before and after the call:
[100047.127123] Init timer 1000
[100048.127986] Expired timer 99790884 99790886
[100049.129578] Expired timer 99791886 99791888
[100050.131146] Expired timer 99792888 99792890
[100051.132728] Expired timer 99793890 99793892
[100052.134315] Expired timer 99794892 99794894
[100053.135882] Expired timer 99795894 99795896
[100054.137411] Expired timer 99796896 99796898
[...]
[100071.164276] Expired timer 99813930 99813932
[100071.529455] Exit timer
And here is the source:
#include <linux/module.h>
#include <linux/kernel.h>
#include <linux/init.h>
#include <linux/jiffies.h>
#include <linux/time.h>
static struct timer_list t;
static void timer_func(unsigned long data)
{
unsigned long pre, post;
t.expires = jiffies + HZ;
pre = t.expires;
add_timer(&t);
post = t.expires;
printk("Expired timer %lu %lu\n", pre, post);
}
static int __init timer_init(void)
{
init_timer(&t);
t.function = timer_func;
t.expires = jiffies + HZ;
add_timer(&t);
printk("Init timer %d\n", HZ);
return 0;
}
static void __exit timer_exit(void)
{
del_timer(&t);
printk("Exit timer\n");
}
module_init(timer_init);
module_exit(timer_exit);
I found the cause. Let's trace the add_timer function:
The add_timer function calls:
mod_timer(timer, timer->expires);
The mod_timer function calls:
expires = apply_slack(timer, expires);
and then goes on to actually modify the timer.
The apply_slack function says:
/*
* Decide where to put the timer while taking the slack into account
*
* Algorithm:
* 1) calculate the maximum (absolute) time
* 2) calculate the highest bit where the expires and new max are different
* 3) use this bit to make a mask
* 4) use the bitmask to round down the maximum time, so that all last
* bits are zeros
*/
Before continuing, let's see what is the timer's slack. The init_timer macro eventually calls do_init_timer which sets the slack by default to -1.
With this knowledge, let's reduce apply_slack and see what remains of it:
static inline
unsigned long apply_slack(struct timer_list *timer, unsigned long expires)
{
unsigned long expires_limit, mask;
int bit;
if (timer->slack >= 0) {
expires_limit = expires + timer->slack;
} else {
long delta = expires - jiffies;
if (delta < 256)
return expires;
expires_limit = expires + delta / 256;
}
mask = expires ^ expires_limit;
if (mask == 0)
return expires;
bit = find_last_bit(&mask, BITS_PER_LONG);
mask = (1 << bit) - 1;
expires_limit = expires_limit & ~(mask);
return expires_limit;
}
The first if, checking for timer->slack >= 0 fails, so the else part is applied. In that part the difference between expires and jiffies is slightly less than HZ (you just did t.expires = jiffies + HZ. Therefore, the delta in the function (with your data) is most likely about 4 and delta / 4 is non zero.
This in turn implies that mask (which is expires ^ expires_limit) is not zero. The rest really depends on the value of expires, but for sure, it gets changed.
So there you have it, since slack is automatically set to -1, the apply_slack function is changing your expires time to align with, I guess, the timer ticks.
If you don't want this slack, you can set t.slack = 0; when you are initializing the timer in timer_init.
This is the old answer! It doesn't address the issue in your question, but it is an issue with what you are trying to achieve nonetheless: having a periodic function.
Let's visualize your program in a timeline (assuming start time 1000 and HZ=50 with imaginary time units):
time (jiffies) event
1000 in timer_init(): t.expires = jiffies + HZ; // t.expires == 1050
1050 timer_func() is called by timer
1052 in timer_func(): t.expires = jiffies + HZ; // t.expires == 1102
1102 timer_func() is called by timer
1104 in timer_func(): t.expires = jiffies + HZ; // t.expires == 1154
I hope you see where this is going! The problem is that there is a delay between the time the timer expires and the time you calculate when the next expiration should be. That's where the drift comes from. The drift could get even larger, by the way, if the system is busy and your function call is delayed.
The way to fix it is very very easy. The problem was that when you update t.expires by jiffies, which is the current time. What you should do is update t.expires by the last time it expired (which is already in t.expires!).
So, in your timer_func function, instead of:
t.expires = jiffies + HZ;
simply do:
t.expires += HZ;
I am trying to use pthread_cond_timedwait for millisecond sleep interval but I am not getting sleep duration. my thread is sleeping more than I have mentioned. below is my implementation. Let me know if i am wrong anywhere.
struct timeval tp;
struct timespec ts;
int rc = gettimeofday(&tp, NULL);
ts.tv_sec = tp.tv_sec;
ts.tv_nsec = tp.tv_usec * 1000;
ts.tv_nsec += 30 * 1000000; //30 is my milliseconds
pthread_mutex_lock(&mtxPlaybackWait);
pthread_cond_timedwait(&playbackSignal, &mtxPlaybackWait, &ts);
pthread_mutex_unlock(&mtxPlaybackWait);
timespac might be overflowed and causing timeout.
Try following:
ts.tv_sec = tp.tv_sec;
ts.tv_nsec = tp.tv_usec * 1000;
ts.tv_nsec += 30 * 1000000;
ts.tv_sec += ts.tv_nsec / 1000000000L;
ts.tv_nsec = ts.tv_nsec % 1000000000L;
You have an addition of seconds and microseconds on one side, and milliseconds on the other. The result is in seconds and nanoseconds.
If you try to express seconds in nanoseconds, this may overflow quickly: 1 second = 1,000,000,000 nanoseconds, which takes up ~30 bits. An unsigned 32-bit integer value can hold up to ~4 seconds if unsigned (~2 for a signed int) and will overflow beyond that.
Also, I am not sure if all functions behave correctly under all circumstances when passed a struct where the fractional seconds amount to more than a second. I’d expect widely used standard libraries to have done their homework and normalize first (or otherwise ensure correct behavior), but some quickly assembled niche product might not handle such cases properly.
To prevent both the overflow and strange side effects of anomalies, shave off integer seconds wherever you can and store them in the seconds part rather than in the fractional seconds.
Here is a version of your calculation which avoids both these things:
gettimeofday(&tp, NULL);
/* if msec is 1 s or more, add its integer part to tv_sec */
ts.tv_sec = tp.tv_sec + floor(msec / 1000);
/* for now, these are really µsec, not nsec, to prevent overflow */
ts.tv_nsec = tp.tv_usec + (msec % 1000) * 1000000;
/* if tv_nsec is 1s or more, move integer second part to tv_sec */
ts.tv_sec += floor(ts.tv_nsec / 1000000);
ts.tv_nsec %= 1000000;
/* and finally, convert µsec to nsec */
ts.tv_nsec *= 1000;
You might not need floor if you are certain that you are operating on integer types (i.e. for msec and ts.tv_nsec)—in that case, a simple division will do.
Does the WaveOut API has some internal limitation of the size for the current piece of buffer played ? I mean if I provide a very small buffer does it affects somehow the sound played to the speakers. I am experiencing very strange noise when I am generating and playing the sinus wave with small buffer. Something like a peak, or "BUMP".
The complete Story:
I made a program that can generate Sinus sound signal in real time.
The variable parameters are Frequency and Volume. The project requirement was to have a maximum latency of 50 ms. So the program must be able to produce Sinus signals with manually adjustable frequency of audio signal in real time.
I used Windows WaveOut API, C# and P/invoke to access the API.
Everything works fine when the sound buffer is 1000 ms large. If I minimize the buffer to 50 ms as per latency requirement then for certain frequencies I am experiencing at the end of every buffer, a noise or "BUMP". I do not understand if the sound generated is malformed ( I checked and is not) or something happens with the Audio chip, or some delay in initializing and playing.
When I save the produced audio to .wav file everything is perfect.
This means the must be some bug in my code or the audio subsystem has a limitation to the buffer chunks sent to it.
For those who doesn't know WaveOut must be initialized at first time and then must be prepared with audio headers for each buffer that are containing the number of bytes that needs to be played and the pointer to a memory that contains the audio that needs to be player.
UPDATE
Noise happens with the following combinations 44100 SamplingRate, 16 Bits, 2 channels, 50 ms buffer and generated Sinus audio signal of 201Hz, 202Hz, 203Hz, 204Hz, 205Hz ... 219Hz,
220Hz, 240 Hz, is ok
Why is this difference of 20, I do not know.
There are a few things to keep in mind when you need to output audio smoothly:
waveOutXxxx API is a legacy/compatibility layer on top of lower level API and as such it has greater overhead and is not recommended when you are to reach minimal latency. Note that this is unlikely to be your primary problem, but this is a piece of general knowledge helpful for understanding
because Windows is not real time OS and its audio subsystem is not realtime either you don't have control over random latency involved between you queue audio data for output and the data is really played back, the key is to keep certain level of buffer fullness which protects you from playback underflows and delivers smooth playback
with waveOutXxxx you are no limited to having single buffer, you can allocate multiple reusable buffers and recycle them
All in all, waveOutXxxx, DirectSound, DirectShow APIs work well with latencies 50 ms and up. With WASAPI exclusive mode streams you can get 5 ms latencies and even lower.
EDIT: I seem to have said too early about 20 ms latencies. To compensate for this, here is a simple tool LowLatencyWaveOutPlay (Win32, x64) to estimate the latency you can achieve. With sufficient buffering playback is smooth, otherwise you hear stuttering.
My understanding is that buffers might be returned late and the optimal design in terms of smallest latency lies along the line of having more smaller buffers so that you are given them back as early as possible. For example, 10 buffers 3 ms/buffer rather than 3 buffers 10 ms/buffer.
D:\>LowLatencyWaveOutPlay.exe 48000 10 3
Format: 48000 Hz, 1 channels, 16 bits per sample
Buffer Count: 10
Buffer Length: 3 ms (288 bytes)
Signal Frequency: 1000 Hz
^C
So I came here because I wanted to find the basic latency of waveoutwrite() as well. I got around 25-26ms of latency before I got to the smooth sine tone.
This is for:
AMD Phenom(tm) 9850 Quad-Core Processor 2.51 GHz
4.00 GB ram
64-bit operating system, x64-based processor
Windows 10 Enterprise N
The code follows. It is a modfied version of Petzold's sine wave program, refactored to run on the command line. I also changed the polling of buffers to use of a callback on buffer complete with the idea that this would make the program more efficient, but it didn't make a difference.
It also has a setup for elapsed timing, which I used to probe various timings for operations on the buffers. Using those I get:
Sine wave output program
Channels: 2
Sample rate: 44100
Bytes per second: 176400
Block align: 4
Bits per sample: 16
Time per buffer: 0.025850
Total time prepare header: 87.5000000000 usec
Total time to fill: 327.9000000000 usec
Total time for waveOutWrite: 90.8000000000 usec
Program:
/*******************************************************************************
WaveOut example program
Based on C. Petzold's sine wave example, outputs a sine wave via the waveOut
API in Win32.
*******************************************************************************/
#include <stdio.h>
#include <windows.h>
#include <math.h>
#include <limits.h>
#include <unistd.h>
#define SAMPLE_RATE 44100
#define FREQ_INIT 440
#define OUT_BUFFER_SIZE 570*4
#define PI 3.14159
#define CHANNELS 2
#define BITS 16
#define MAXTIM 1000000000
double fAngle;
LARGE_INTEGER perffreq;
PWAVEHDR pWaveHdr1, pWaveHdr2;
int iFreq = FREQ_INIT;
VOID FillBuffer (short* pBuffer, int iFreq)
{
int i;
int c;
for (i = 0 ; i < OUT_BUFFER_SIZE ; i += CHANNELS) {
for (c = 0; c < CHANNELS; c++)
pBuffer[i+c] = (short)(SHRT_MAX*sin (fAngle));
fAngle += 2*PI*iFreq/SAMPLE_RATE;
if (fAngle > 2 * PI) fAngle -= 2*PI;
}
}
double elapsed(LARGE_INTEGER t)
{
LARGE_INTEGER rt;
long tt;
QueryPerformanceCounter(&rt);
tt = rt.QuadPart-t.QuadPart;
return (tt*(1.0/(double)perffreq.QuadPart));
}
void CALLBACK waveOutProc(HWAVEOUT hwo, UINT uMsg, DWORD_PTR dwInstance, DWORD_PTR dwParam1, DWORD_PTR dwParam2)
{
if (uMsg == WOM_DONE) {
if (pWaveHdr1->dwFlags & WHDR_DONE) {
FillBuffer((short*)pWaveHdr1->lpData, iFreq);
waveOutWrite(hwo, pWaveHdr1, sizeof(WAVEHDR));
}
if (pWaveHdr2->dwFlags & WHDR_DONE) {
FillBuffer((short*)pWaveHdr2->lpData, iFreq);
waveOutWrite(hwo, pWaveHdr2, sizeof(WAVEHDR));
}
}
}
int main()
{
HWAVEOUT hWaveOut ;
short* pBuffer1;
short* pBuffer2;
short* pBuffer3;
WAVEFORMATEX waveformat;
UINT wReturn;
int bytes;
long t;
LARGE_INTEGER rt;
double timprep;
double filtim;
double waveouttim;
printf("Sine wave output program\n");
fAngle = 0; /* start sine angle */
QueryPerformanceFrequency(&perffreq);
pWaveHdr1 = malloc (sizeof (WAVEHDR));
pWaveHdr2 = malloc (sizeof (WAVEHDR));
pBuffer1 = malloc (OUT_BUFFER_SIZE*sizeof(short));
pBuffer2 = malloc (OUT_BUFFER_SIZE*sizeof(short));
pBuffer3 = malloc (OUT_BUFFER_SIZE*sizeof(short));
if (!pWaveHdr1 || !pWaveHdr2 || !pBuffer1 || !pBuffer2) {
if (!pWaveHdr1) free (pWaveHdr1) ;
if (!pWaveHdr2) free (pWaveHdr2) ;
if (!pBuffer1) free (pBuffer1) ;
if (!pBuffer2) free (pBuffer2) ;
fprintf(stderr, "*** Error: No memory\n");
exit(1);
}
// Load prime parameters to format
waveformat.wFormatTag = WAVE_FORMAT_PCM;
waveformat.nChannels = CHANNELS;
waveformat.nSamplesPerSec = SAMPLE_RATE;
waveformat.wBitsPerSample = BITS;
waveformat.cbSize = 0;
// Calculate other parameters
bytes = waveformat.wBitsPerSample/8; /* find bytes per sample */
if (waveformat.wBitsPerSample&8) bytes++; /* round up */
bytes *= waveformat.nChannels; /* find total channels size */
waveformat.nBlockAlign = bytes; /* set block align */
/* find average bytes/sec */
waveformat.nAvgBytesPerSec = bytes*waveformat.nSamplesPerSec;
printf("Channels: %d\n", waveformat.nChannels);
printf("Sample rate: %d\n", waveformat.nSamplesPerSec);
printf("Bytes per second: %d\n", waveformat.nAvgBytesPerSec);
printf("Block align: %d\n", waveformat.nBlockAlign);
printf("Bits per sample: %d\n", waveformat.wBitsPerSample);
printf("Time per buffer: %f\n",
OUT_BUFFER_SIZE*sizeof(short)/(double)waveformat.nAvgBytesPerSec);
if (waveOutOpen (&hWaveOut, WAVE_MAPPER, &waveformat, (DWORD_PTR)waveOutProc, 0, CALLBACK_FUNCTION)
!= MMSYSERR_NOERROR) {
free (pWaveHdr1) ;
free (pWaveHdr2) ;
free (pBuffer1) ;
free (pBuffer2) ;
hWaveOut = NULL ;
fprintf(stderr, "*** Error: No memory\n");
exit(1);
}
// Set up headers and prepare them
pWaveHdr1->lpData = (LPSTR)pBuffer1;
pWaveHdr1->dwBufferLength = OUT_BUFFER_SIZE*sizeof(short);
pWaveHdr1->dwBytesRecorded = 0;
pWaveHdr1->dwUser = 0;
pWaveHdr1->dwFlags = WHDR_DONE;
pWaveHdr1->dwLoops = 1;
pWaveHdr1->lpNext = NULL;
pWaveHdr1->reserved = 0;
QueryPerformanceCounter(&rt);
waveOutPrepareHeader(hWaveOut, pWaveHdr1, sizeof (WAVEHDR));
timprep = elapsed(rt);
pWaveHdr2->lpData = (LPSTR)pBuffer2;
pWaveHdr2->dwBufferLength = OUT_BUFFER_SIZE*sizeof(short);
pWaveHdr2->dwBytesRecorded = 0;
pWaveHdr2->dwUser = 0;
pWaveHdr2->dwFlags = WHDR_DONE;
pWaveHdr2->dwLoops = 1;
pWaveHdr2->lpNext = NULL;
pWaveHdr2->reserved = 0;
waveOutPrepareHeader(hWaveOut, pWaveHdr2, sizeof (WAVEHDR));
// Send two buffers to waveform output device
QueryPerformanceCounter(&rt);
FillBuffer (pBuffer1, iFreq);
filtim = elapsed(rt);
QueryPerformanceCounter(&rt);
waveOutWrite (hWaveOut, pWaveHdr1, sizeof (WAVEHDR));
waveouttim = elapsed(rt);
FillBuffer (pBuffer2, iFreq);
waveOutWrite (hWaveOut, pWaveHdr2, sizeof (WAVEHDR));
// Run waveform loop
sleep(10);
printf("Total time prepare header: %.10f usec\n", timprep*1000000);
printf("Total time to fill: %.10f usec\n", filtim*1000000);
printf("Total time for waveOutWrite: %.10f usec\n", waveouttim*1000000);
waveOutUnprepareHeader(hWaveOut, pWaveHdr1, sizeof (WAVEHDR));
waveOutUnprepareHeader(hWaveOut, pWaveHdr2, sizeof (WAVEHDR));
// Close waveform file
free (pWaveHdr1) ;
free (pWaveHdr2) ;
free (pBuffer1) ;
free (pBuffer2) ;
}
There are a number of posts and references on how to get CPU Utilization using statistics in /proc/stat. However, most of them use only four of the 7+ CPU stats (user, nice, system, and idle), ignoring the remaining jiffie CPU counts present in Linux 2.6 (iowait, irq, softirq).
As an example, see Determining CPU utilization.
My question is this: Are the iowait/irq/softirq numbers also counted in one of the first four numbers (user/nice/system/idle)? In other words, does the total jiffie count equal the sum of the first four stats? Or, is the total jiffie count equal to the sum of all 7 stats? If the latter is true, then a CPU utilization formula should take all of the numbers into account, like this:
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
long double a[7],b[7],loadavg;
FILE *fp;
for(;;)
{
fp = fopen("/proc/stat","r");
fscanf(fp,"%*s %Lf %Lf %Lf %Lf",&a[0],&a[1],&a[2],&a[3],&a[4],&a[5],&a[6]);
fclose(fp);
sleep(1);
fp = fopen("/proc/stat","r");
fscanf(fp,"%*s %Lf %Lf %Lf %Lf",&b[0],&b[1],&b[2],&b[3],&b[4],&b[5],&b[6]);
fclose(fp);
loadavg = ((b[0]+b[1]+b[2]+b[4]+b[5]+b[6]) - (a[0]+a[1]+a[2]+a[4]+a[5]+a[6]))
/ ((b[0]+b[1]+b[2]+b[3]+b[4]+b[5]+b[6]) - (a[0]+a[1]+a[2]+a[3]+a[4]+a[5]+a[6]));
printf("The current CPU utilization is : %Lf\n",loadavg);
}
return(0);
}
I think iowait/irq/softirq are not counted in one of the first 4 numbers. You can see the comment of irqtime_account_process_tick in kernel code for more detail:
(for Linux kernel 4.1.1)
2815 * Tick demultiplexing follows the order
2816 * - pending hardirq update <-- this is irq
2817 * - pending softirq update <-- this is softirq
2818 * - user_time
2819 * - idle_time <-- iowait is included in here, discuss below
2820 * - system time
2821 * - check for guest_time
2822 * - else account as system_time
For the idle time handling, see account_idle_time function:
2772 /*
2773 * Account for idle time.
2774 * #cputime: the cpu time spent in idle wait
2775 */
2776 void account_idle_time(cputime_t cputime)
2777 {
2778 u64 *cpustat = kcpustat_this_cpu->cpustat;
2779 struct rq *rq = this_rq();
2780
2781 if (atomic_read(&rq->nr_iowait) > 0)
2782 cpustat[CPUTIME_IOWAIT] += (__force u64) cputime;
2783 else
2784 cpustat[CPUTIME_IDLE] += (__force u64) cputime;
2785 }
If the cpu is idle AND there is some IO pending, it will count the time in CPUTIME_IOWAIT. Otherwise, it is count in CPUTIME_IDLE.
To conclude, I think the jiffies in irq/softirq should be counted as "busy" for cpu because it was actually handling some IRQ or soft IRQ. On the other hand, the jiffies in "iowait" should be counted as "idle" for cpu because it was not doing something but waiting for a pending IO to happen.
from busybox, its top magic is:
static const char fmt[] ALIGN1 = "cp%*s %llu %llu %llu %llu %llu %llu %llu %llu";
int ret;
if (!fgets(line_buf, LINE_BUF_SIZE, fp) || line_buf[0] != 'c' /* not "cpu" */)
return 0;
ret = sscanf(line_buf, fmt,
&p_jif->usr, &p_jif->nic, &p_jif->sys, &p_jif->idle,
&p_jif->iowait, &p_jif->irq, &p_jif->softirq,
&p_jif->steal);
if (ret >= 4) {
p_jif->total = p_jif->usr + p_jif->nic + p_jif->sys + p_jif->idle
+ p_jif->iowait + p_jif->irq + p_jif->softirq + p_jif->steal;
/* procps 2.x does not count iowait as busy time */
p_jif->busy = p_jif->total - p_jif->idle - p_jif->iowait;
}
I've got a piece of userspace code which is parsing /proc/PID/task/TID/stat to get the cpu usage. I can use HZ to get the jiffies per second but this code could move to another machine which has a different configured value. Is there any way to get the value of HZ from userspace at runtime?
You divide it by the number you get from sysconf(_SC_CLK_TCK).
However, I think this is probably always 100 under Linux regardless of the actual clock tick, it's always presented to userspace as 100.
See man proc(5).
To clarify the math behind MarkR's answer:
sysconf(_SC_CLK_TCK) will get you jiffies per second. Divide jiffies by the number you get from sysconf(_SC_CLK_TCK) to get the total number of seconds.
jiffies jiffies seconds
-------------------- = ----------------- = ------- = seconds
sysconf(_SC_CLK_TCK) (jiffies/second) 1
Source of "ps" command include file <linux/param.h> to get value of HZ.
They also look for an "ELF note" with number 17 to find value of HZ (sysinfo.c):
//extern char** environ;
/* for ELF executables, notes are pushed before environment and args */
static unsigned long find_elf_note(unsigned long findme){
unsigned long *ep = (unsigned long *)environ;
while(*ep++);
while(*ep){
if(ep[0]==findme) return ep[1];
ep+=2;
}
return NOTE_NOT_FOUND;
}
[...]
hz = find_elf_note(17);
I have to admit it look weird for me since ELF notes is a section defined during compilation.
For shell-scripting, etc, use getconf CLK_TCK from the command-line. Use can use this to pass that parameter in as an environment variable or on the command-line.
main(int argc, char **argv) {
unsigned long clk_tck = atol(
getenv("CLK_TCK") || "0"
) || sysconf(_SC_CLK_TCK) ;
... /* your code */
This uses the sysconf as above, but allows you to override it with an environment variable, which can be set with the above command.