linux - nanosleep sleep 60 微秒太长

Question

我在 g++ 中编译了以下测试，nanosleep 太长了，它需要 60 微秒完成，我预计它只花费不到 1 微秒:

int main()
{
    gettimeofday(&startx, NULL);
    struct timespec req={0};
    req.tv_sec=0;
    req.tv_nsec=100 ;
    nanosleep(&req,NULL) ;
    gettimeofday(&endx, NULL);
    printf("(%d)(%d)\n",startx.tv_sec,startx.tv_usec);
    printf("(%d)(%d)\n",endx.tv_sec,endx.tv_usec);
    return 0 ;
}

我的环境:uname -r 显示:

3.10.0-123.el7.x86_64

cat/boot/config-uname -r | grep 赫兹

CONFIG_NO_HZ_COMMON=y
# CONFIG_HZ_PERIODIC is not set
# CONFIG_NO_HZ_IDLE is not set
CONFIG_NO_HZ_FULL=y
# CONFIG_NO_HZ_FULL_ALL is not set
CONFIG_NO_HZ=y
# CONFIG_RCU_FAST_NO_HZ is not set
# CONFIG_HZ_100 is not set
# CONFIG_HZ_250 is not set
# CONFIG_HZ_300 is not set
CONFIG_HZ_1000=y
CONFIG_HZ=1000
CONFIG_MACHZ_WDT=m

我是否应该在 HZ 配置中做一些事情，以便 nanosleep 完全符合我的预期？!

我的CPU信息:

Architecture:          x86_64
CPU op-mode(s):        32-bit, 64-bit
Byte Order:            Little Endian
CPU(s):                24
On-line CPU(s) list:   0-23
Thread(s) per core:    2
Core(s) per socket:    6
Socket(s):             2
NUMA node(s):          2
Vendor ID:             GenuineIntel
CPU family:            6
Model:                 63
Model name:            Intel(R) Xeon(R) CPU E5-2643 v3 @ 3.40GHz
Stepping:              2
CPU MHz:               3600.015
BogoMIPS:              6804.22
Virtualization:        VT-x
L1d cache:             32K
L1i cache:             32K
L2 cache:              256K
L3 cache:              20480K
NUMA node0 CPU(s):     0,2,4,6,8,10,12,14,16,18,20,22
NUMA node1 CPU(s):     1,3,5,7,9,11,13,15,17,19,21,23

编辑:

#ifdef SYS_gettid
    pid_t tid = syscall(SYS_gettid);
    printf("thread1...tid=(%d)\n",tid);
#else
    #error "SYS_gettid unavailable on this system"
#endif

这将获取我喜欢的高优先级线程的消息，然后执行chrt -v -r -p 99 tid 实现这个目标，感谢 Johan Boule亲切的帮助，非常感谢!!!

编辑2:

#ifdef SYS_gettid
    const char *sched_policy[] = {
    "SCHED_OTHER",
    "SCHED_FIFO",
    "SCHED_RR",
    "SCHED_BATCH"
    };
    struct sched_param sp = {
        .sched_priority = 99
    };
    pid_t tid = syscall(SYS_gettid);
    printf("thread1...tid=(%d)\n",tid);
    sched_setscheduler(tid, SCHED_RR, &sp);
    printf("Scheduler Policy is %s.\n", sched_policy[sched_getscheduler(0)]);
#else
    #error "SYS_gettid unavailable on this system"
#endif

这将在没有 chrt 帮助的情况下完成我想做的事情

Answer 1

(这不是答案)

值得一提的是，使用更现代的函数会导致相同的结果:

#include <stddef.h>
#include <time.h>
#include <stdio.h>

int main()
{
    struct timespec startx, endx;
    clock_gettime(CLOCK_MONOTONIC, &startx);
    struct timespec req={0};
    req.tv_sec=0;
    req.tv_nsec=100 ;
    clock_nanosleep(CLOCK_MONOTONIC, 0, &req, NULL);
    clock_gettime(CLOCK_MONOTONIC, &endx);
    printf("(%d)(%d)\n",startx.tv_sec,startx.tv_nsec);
    printf("(%d)(%d)\n",endx.tv_sec,endx.tv_nsec);
    return 0 ;
}

输出:

(296441)(153832940)
(296441)(153888488)