任务调度与上下文切换时间测试

创建两个进程(实时进程)并在它们之间传送一个令牌，如此往返传送一定的次数。其中一个进程在读取令牌时就会引起阻塞。另一个进程发送令牌后等待其返回时也处于阻塞状态。发送令牌带来的开销与上下文切换带来的开销相比，可以忽略不计（经测试，一次管道传递约用时3ns左右）。 (利用管道传递令牌)

方法一：使用gettimeofday()

//方法一：使用gettimeofday()获取当前时间

#include <stdio.h>
#include <stdlib.h>
#include <sys/time.h>
#include <time.h>
#include <sched.h>
#include <sys/types.h>
#include <unistd.h>      //pipe()

int main()
{
int x, i, fd[2], p[2];
int tv[2];	//用于进程间数据通信
char send    = 's';
char receive;
pipe(fd);
pipe(p);
pipe(tv);
struct timeval tv_start,tv_end;
struct sched_param param;
param.sched_priority = 0;

while ((x = fork()) == -1);
if (x==0) {			//子进程
sched_setscheduler(getpid(), SCHED_FIFO, ¶m);
gettimeofday(&tv_start, NULL);
write(tv[1], &tv_start, sizeof(tv_start));//保存数据，以便传递到父进程中进行计算
//printf("Before Context Switch Time1.sec %u s\n", tv_start.tv_sec);
//printf("Before Context Switch Time1.usec %u us\n", tv_start.tv_usec);
for (i = 0; i < 10000; i++) {
read(fd[0], &receive, 1);
//printf("Child read!\n");
write(p[1], &send, 1);
//printf("Child write!\n");
}
exit(0);
}
else {		//父进程
sched_setscheduler(getpid(), SCHED_FIFO, ¶m);
for (i = 0; i < 10000; i++) {
write(fd[1], &send, 1);
//printf("Parent write!\n");
read(p[0], &receive, 1);
//printf("Parent read!\n");
}
gettimeofday(&tv_end, NULL);
//printf("After Context SWitch Time1.sec %u s\n", tv_end.tv_sec);
//printf("After Context SWitch Time1.usec %u us\n", tv_end.tv_usec);

}
read(tv[0], &tv_start, sizeof(tv_start));
//printf("Before Context Switch Time2.sec %u s\n", tv_start.tv_sec);
//printf("Before Context Switch Time2.usec %u us\n", tv_start.tv_usec);
//printf("Before Context Switch Time %u us\n", tv_start.tv_usec);
//printf("After Context SWitch Time2.sec %u s\n", tv_end.tv_sec);
//printf("After Context SWitch Time2.usec %u us\n", tv_end.tv_usec);
printf("Task Switch Time: %f us\n", (1000000*(tv_end.tv_sec-tv_start.tv_sec) + tv_end.tv_usec - tv_start.tv_usec)/20000.0);
return 0;
}