我测试了一个程序,以计算a中的计算时间 机器有48个intel cpu核心。
1)运行./test时,输出如下:
MM: 324.039000 ms
MM: 324.052000 ms
MM: 324.079000 ms
MM: 324.042000 ms
MM: 324.060000 ms
MM: 324.052000 ms
MM: 324.026000 ms
MM: 324.040000 ms
MM: 324.075000 ms
MM: 324.011000 ms
MM: 324.072000 ms
2)运行其他程序,例如示例代码2
./aa &
./aa &
....
about 30 times
3)./ test输出更改:
MM: 354.601000 ms
MM: 322.289000 ms
MM: 322.250000 ms
MM: 322.348000 ms
MM: 354.700000 ms
MM: 322.556000 ms
MM: 354.834000 ms
MM: 354.816000 ms
MM: 354.777000 ms
MM: 354.812000 ms
MM: 322.451000 ms
MM: 322.507000 ms
MM: 322.473000 ms
4)为了减少计算成本时间,我尝试过:
SCHED_RR
/ SCHED_FIFO
,优先级为50/98
chrt -f -p 98 5911
taskset -pc 4 5911
for i in $(ps -ef |grep aa|awk '{print $2}'); do echo $i; taskset -pc 5-47 $i; done
它仍然不起作用,结果不好。
MM: 322.316000 ms
MM: 322.307000 ms
MM: 326.739000 ms
MM: 354.399000 ms
MM: 354.662000 ms
MM: 354.740000 ms
MM: 322.275000 ms
for i in $(ps -ef |grep aa|awk '{print $2}'); do echo $i; taskset -pc 5-7 $i; done
结果:
MM: 322.304000 ms
MM: 322.335000 ms
MM: 322.344000 ms
MM: 322.319000 ms
MM: 322.340000 ms
MM: 322.328000 ms
MM: 322.298000 ms
MM: 322.318000 ms
为什么这种低优先级进程会影响实时性,这真是太神奇了 过程,请帮帮我!
示例测试代码1:
#include <stdio.h>
#include <sys/time.h>
void calc() {
int c, d, k;
int m = 100000, q = 1, p = 1000;
int sum;
for (c = 0; c < m; c++) {
for (d = 0; d < q; d++) {
for (k = 0; k < p; k++) {
sum = c * m * d * q * p * k;
}
}
}
}
int main() {
struct timeval t1, t2;
double elapsedTime;
while(1) {
gettimeofday(&t1, NULL);
calc();
gettimeofday(&t2, NULL);
elapsedTime = (t2.tv_sec - t1.tv_sec) * 1000.0; // sec to ms
elapsedTime += (t2.tv_usec - t1.tv_usec) / 1000.0; // us to ms
printf("MM: %lf ms\n", elapsedTime);
sleep(1);
}
}
示例代码2:
#include <stdio.h>
int main() {
while(1) {
}
}