[PATCH] makedumpfile: print spinner in progress information

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(2013/10/25 13:07), Atsushi Kumagai wrote:
> Hello HATAYAMA-san,
> 
> (2013/10/25 9:55), HATAYAMA Daisuke wrote:
>> On system with huge memory, percentage in progress information is
>> updated at very slow interval, because 1 percent on 1 TiB memory is
>> about 10 GiB, which looks like as if system has freezed. Then,
>> confused users might get tempted to push a reset button to recover the
>> system. We want to avoid such situation as much as possible.
>>
>> To address the issue, this patch adds spinner that rotates in the
>> order of /, |, \ and - next to the progress indicator in percentage,
>> which helps users to get aware that system is still active and crash
>> dump process is still in progress now.
>>
>> This code is borrowed from diskdump code.
>>
>> The example is like this:
>>
>> Copying data                       : [  0 %] /
>> Copying data                       : [  8 %] |
>> Copying data                       : [ 11 %] \
>> Copying data                       : [ 14 %] -
>> Copying data                       : [ 16 %] /
>> ...
>> Copying data                       : [ 99 %] /
>> Copying data                       : [100 %] |
> 
> I like it, but have a comment.
> 
>      6109 int
>      6110 write_kdump_pages_cyclic(struct cache_data *cd_header, struct cache_data *cd_page,
>      6111                          struct page_desc *pd_zero, off_t *offset_data)
>      6112 {
>      ...
>      6156         per = info->num_dumpable / 100;
>      ...
>      6178         for (pfn = start_pfn; pfn < end_pfn; pfn++) {
>      6179
>      6180                 if ((num_dumped % per) == 0)
>      6181                         print_progress(PROGRESS_COPY, num_dumped, info->num_dumpable);
> 
> The interval of calling print_progress() looks still long if
> num_dumpable is huge.
> So how about fix this, e.g., by changing the interval to time based ?
> 

I wrote simple bench for time-based interval as below, which measures
total time consumed for calling time system call with/without vDSO.
It seems to me that both results are acceptable.
I'll reflect this change in next version.

$ ./bench
total: 21.059131
average: 0.000000
total: 65.558263
average: 0.000000

==bench.c
#include <stdio.h>
#include <stdlib.h>
#include <time.h>

static inline double getdtime(void)
{
        struct timeval tv;
        gettimeofday(&tv, NULL);
        return (double)tv.tv_sec + (double)tv.tv_usec * 1.0e-6;
}

int main(int argc, char **argv)
{
        unsigned long i;
        time_t t;
        double t1, t2, total;
        const int NR_time = 201;
        const unsigned long nr_repeat = (1UL << 40) / 4096;

        for (i = 0; i < nr_repeat; ++i) {
                t1 = getdtime();
                time(&t);
                t2 = getdtime();
                total += t2 - t1;
        }
        printf("total: %lf\n", total);
        printf("average: %lf\n", total / nr_repeat);

        for (i = 0; i < nr_repeat; ++i) {
                t1 = getdtime();
                syscall(NR_time, &t);
                t2 = getdtime();
                total += t2 - t1;
        }
        printf("total: %lf\n", total);
        printf("average: %lf\n", total / nr_repeat);

        return 0;
}
==

-- 
Thanks.
HATAYAMA, Daisuke




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